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bạn có thể cho tớ câu trả lời rõ àng hơn không

a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3

x : 0,1 = 12

x x 10 = 12

x = 12 : 10

x = 1,2

--------------------------------------------

4 : 0,25 + 320 x 0,5 + 3: 0,125

= 4 x 4 + 320 : 2 + 3 x 8

= 16 + 160 + 24

= 160 + (16 + 24)

= 160 + 40

= 200

rút gọn à banj

đúng rồi á

a/

\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+66.67+66.2=\)

\(=\left(1.2+2.3+3.4+...+66.67\right)+2\left(1+2+3+...+66\right)\)

Đặt

\(B=1+2+3+...+66=\dfrac{66\left(1+66\right)}{2}=2211\)

Đặt 

\(C=1.2+2.3+3.4+...+66.67\)

\(3C=1.2.3+2.3.3+3.4.3+...+66.67.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+66.67.\left(68-65\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-65.66.67+66.67.68=\)

\(=66.67.68\Rightarrow C=\dfrac{66.67.68}{3}=22.67.68\)

\(\Rightarrow A=C+2B\) Bạn tự tính nhé

b/

\(B=2\left(1.50+2.49+3.48+...+25.26\right)=\)

Ta có

\(C=1.50+2.49+3.48+...+25.26=\)

\(\left(50-49\right).50+\left(50-48\right).49+\left(50-47\right).48+...+\left(50-25\right).26=\)

\(=50.50-49.50+50.49-48.49+50.48-47.48+50.26-25.26=\)

\(=50.\left(26+27+28+...+50\right)-\left(25.26+26.27+27.28+...+49.50\right)\)

Ta có

\(D=26+27+28+...+50=\dfrac{25.\left(26+50\right)}{2}=950\)

Ta có

\(E=25.26+26.27+27.28+...+49.50\)

\(3E=25.26.3+26.27.3+27.28.3+...+49.50.3=\)

\(=25.26.\left(27-24\right)+26.27.\left(28-25\right)+...+49.50.\left(51-48\right)=\)

\(=-24.25.26+25.26.27-25.26.27+26.27.28-...-48.49.50+49.50.51=\)

\(=49.50.51-24.25.26\)

\(\Rightarrow E=\dfrac{49.50.51-24.25.26}{3}\)

\(\Rightarrow C=50D-E\)

\(B=2C\)

Bạn tự tính nhé

\(\left|3x+2\right|=\left|4x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x-3\\3x+2=3-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2+3x=4x-3\\2+3x=3-4x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

a) Đặt  \(A=4x-x^2-5\)

\(-A=x^2-4x+5\)

\(-A=\left(x^2-4x+4\right)+1\)

\(-A=\left(x-2\right)^2+1\)

Mà  \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge1\)

\(\Leftrightarrow A\le-1< 0\left(đpcm\right)\)

b) Đặt  \(B=x^2-2x+5\)

\(B=\left(x^2-2x+1\right)+4\)

\(B=\left(x-1\right)^2+4\)

Mà  \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\left(đpcm\right)\)

a)4x-x²-5 = -(x²-4x+4)-1= -(x-2)^2 -1 < 0 với mọi x (đpcm)

b) x² -2x+5= (x²-2x+1)+4=(x-1)^2 +4 >0  với mọi x (đpcm)

Thế S là số nào bn mà chia hết cho 3 vậy bn ?