i) x3- 11x2 + 30x;
j) 4x4- 21x2y2 + y4
k) x3 + 4x2- 7x - 10;
l) (x2 + x)2- (x2 + x) + 15;
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a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
Đáp án A
Từ phương trình ta suy ra
x 4 − x 3 + 50 x 2 − 60 x + 20 = 13 x 3 − 11 x 2 + 22 x − 2 ⇔ x 4 − 14 x 3 + 61 x 2 − 82 x + 22 = 0 ⇔ x 4 − 8 x + 11 x 2 − 6 x + 2 = 0 ⇔ x = 3 − 7 x = 4 − 5 x = 3 + 7 x = 4 + 5
Ta đã biết phương trình đã cho có 4 nghiệm nên ta có
a = 3 − 7 ; c = 3 + 7 .
Do đó P = a 2 + c 2 = 32.
a: \(\Leftrightarrow8x^2+16x+14x+7=0\)
=>(2x+1)(8x+7)=0
=>x=-1/2 hoặc x=-7/8
b: \(=x^3-x-6x-6\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
a) \(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
b) \(=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)
c) \(=3x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(3x+5\right)\)
d) \(=\left(x+y\right)^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
e) \(=x\left(x^2-11x+30\right)\)
f) \(=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)
a) 4,25 x 58,47 + 41,53 x 4,25
= 4,25 x ( 58,47 + 41,53 )
= 4,25 x 100
= 425
nhớ k đúng cho mik nha
bài này dễ thui
\(55555:11.2\)
\(=\frac{55555}{11}.2\)
\(=10100.90909\)
mk nghĩ zầy đó
ahjhj
Huynh Mai Thao
<=>\(x^4-8x^2-9=0\)
đặt \(t=x^2\left(t\ge0\right)\)
pt thành \(t^2-8t-9=0\)
<=>\(\left(t+1\right)\left(t-9\right)\)
<=>\(\left[{}\begin{matrix}t=-1\\t=9\end{matrix}\right.\)
so với đk của t=>t=9
vậy \(x^2=9\)
<=>x=\(\pm9\)
i) x3- 11x2 + 30x
=\(x\left(x^2-11x+30\right)\)
=\(x\left(x-6\right)\left(x-5\right)\)
j) 4x4- 21x2y2 + y4
=4x^4+4x^2y^2+y^4-25x^2y^2
=(2x^2+y^2)^2-(5xy)^2
=(2x^2+y^2-5xy)(2x^2+y^2+5xy)