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a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
1, x(x-1)=2(x-1)
<=> x(x-1)-2(x-1)=0
<=> (x-2)(x-1)=0
<=>x=2 hoặc x=1
vậy ...
2, (x+2)(2x-3)=x^2 -4
<=>(x+2)(2x-3)=(x-2)(x+2)
<=> (x+2)(2x-3)-(x-2)(x+2)=0
<=> (x+2)(2x-3-x+2)=0
<=> x=-2 hoặc x=1
vây...
3,x^2 +3x +2=0
<=> x^2 +x+2x+2=0
<=>(x+2)(x+1)=0
<=> x=-2 hoặc x=-1
vậy ...
5, x^3+x^2-12x =0
<=> x(x^2+x-12)=0
<=>x(x^2-3x+4x-12)=0
<=>x(x+4)(x-3)=0
<=> x=0 hoặc x=-4 hoặc x=3
vậy ...
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
chẳng ai giải, thôi mình giải vậy!
a) Đặt \(y=x^2+4x+8\),phương trình có dạng:
\(t^2+3x\cdot t+2x^2=0\)
\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)
\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}
b) nhân 2 vế của phương trình với 12 ta được:
\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)
Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)
giải tiếp ra ta sẽ được S={-2/3;-5/3}
c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)
S={3;5}
d)s={1}
e) S={1;-2;-1/2}
f) phương trình vô nghiệm
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
a. 7(2x - 0,5) - 3(x + 4) = 4 - 5(x - 0,7)
⇔ 14x - 4,5 - 3x - 12 = 4 - 5x + 3,5
⇔ 14x -3x + 5x = 4 + 4,5 + 3,5
⇔ 16x = 12
⇔ x = \(\dfrac{12}{16}=\dfrac{3}{4}\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
Tớ học ngu nên chỉ biết cách nhân ra rồi rút gọn chứ không biết cách nào ngắn hơn :)) Hơi dài dòng nên phân tích từng vế 1 nhé :D
2/ \(\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)=4\left(2x^2+5x-204\right)\left(x^2-5x-206\right)\)
*****\(VT=\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)^2\)
\(=4x^4+25x^2+41616+20x^3-816x^2-2040x+4\left(x^4-387x^2+42436-10x^3+2060x\right)\)
\(=4x^2+25x^2+41616+20x^3-816x^2-2040x+4x^2-1548x^2+169744-40x^3+8240x\)
\(=8x^4-1523x^2+6200x+211360\)
*****\(VP=\left(8x^2+20x-816\right)\left(x^2-5x-206\right)\)
\(=8x^4-40x^3-1648x^2-100x^2-4120x-816x^2+4080x+168096\)
\(=8x^4-1748x^2-40x+168096\)
\(\Rightarrow8x^4-1523x^2+6200x+211360=8x^4-1748x^2-40x+168096\)
\(\Leftrightarrow-1523x^2+6200x+211360+1748x^2-40x+168096=0\)
\(\Leftrightarrow255x^2+43264+6240x=0\)
\(\Leftrightarrow\left(15x+208\right)^2=0\)
\(\Leftrightarrow15x+208=0\)
\(\Leftrightarrow x=-\frac{208}{15}\)
+ Ta có: \(x^4-5x^3+6x^2+5x+1=0\)
\(\Rightarrow x^2-5x+6+\frac{5}{x}+\frac{1}{x^2}=0\)( chia cả hai vế cho \(x^2\))
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-\left(5x-\frac{5}{x}\right)+6=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-5.\left(x-\frac{1}{x}\right)+6=0\)( *** )
- Đặt \(x-\frac{1}{x}=a\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=a^2+2\)
- Thay \(a=x-\frac{1}{x};\)\(a^2+2=x^2+\frac{1}{x^2}\)vào ( *** )
- Ta có: \(a^2+2-5a+6=0\)
\(\Leftrightarrow a^2-5a+8=0\)
\(\Leftrightarrow4a^2-20a+32=0\)
\(\Leftrightarrow\left(4a^2-20a+25\right)+7=0\)
\(\Leftrightarrow\left(2a-5\right)^2+7=0\)
- Ta lại có: \(\hept{\begin{cases}\left(2a-5\right)^2\ge0\forall a\\7>0\end{cases}}\Rightarrow \left(2a-5\right)^2+7\ge7>0\)mà \(\left(2a-5\right)^2+7=0\)
\(\Rightarrow\left(2a-5\right)^2+7\)( vô nghiệm ) \(\Rightarrow\)\(x^4-5x^3+6x^2+5x+1=0\)( vô nghiệm )
Vậy \(S=\left\{\varnothing\right\}\)
+ Ta có: \(\left(2x^2+5x-204\right)^2+4.\left(x^2-5x-206\right)=4.\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)( ** )
- Đặt \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)
- Thay \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)
vào ( ** )
- Ta có: \(a^2+4b^2=4ab\)
\(\Leftrightarrow a^2-4ab+4b^2=0\)
\(\Leftrightarrow\left(a-2b\right)^2=0\)
\(\Leftrightarrow a-2b=0\)
\(\Leftrightarrow a=2b\)( * )
- Thay \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)vào ( * )
- Ta lại có: \(2x^2+5x-204=2.\left(x^2-5x-206\right)\)
\(\Leftrightarrow2x^2+5x-204=2x^2-10x-412\)
\(\Leftrightarrow\left(2x^2-2x^2\right)+\left(5x+10x\right)=-\left(412-204\right)\)
\(\Leftrightarrow15x=-208\)
\(\Leftrightarrow x=-\frac{208}{15} \left(TM\right)\)
Vậy \(S=\left\{-\frac{208}{15}\right\}\)