a (3-2x)2 = 6 - 4x

b (xy+5)2 = 2xy + 10

c (2x+1)(1-2x) = 2x - 4x² + 1 - 2x = 4x2 + 1

d (1-5x)3 = 3-15x

e (2x+y)(4x2 - 4xy + y2) = 8x³ -8x²y+2xy² + 4x²y-4xy² + y³ = 8x³ + y³ - 4x²y - 2xy²

\(M=343-8x^3-64+8x^3=279\\ N=8x^3-1-1+8x^3=16x^3=16\cdot1000=16000\)

\(M=343-8x^3-\left(64-8x^3\right)=343-64=279\)

(biểu thức này ko phụ thuộc vào biến x nên ko cần thay)

\(N=8x^3-1-1+8x^3=16x^3\)

Thay \(x=10\)

\(N=16\cdot10^3=16\cdot1000=16000\)

\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)

a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)

\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)

\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)

b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)

\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)

\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)

\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)

c, \(4x^2-2x-1=0\)

\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)

\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)

\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)

d,\(x^4-4x^2-32=0\)

đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)

\(< =>t^2-2.2t+4-6^2=0\)

\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)

\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)

\(2x+\dfrac{3}{7}=4x-\dfrac{1}{15}\)

\(\Leftrightarrow-2x=-\dfrac{1}{15}-\dfrac{3}{7}\)

\(\Leftrightarrow-2x=-\dfrac{52}{105}\)

\(\Leftrightarrow x=\dfrac{26}{105}\)

Mk thấy cái này có vẻ đúng hơn :)

\(\dfrac{2x+3}{7}=\dfrac{4x-1}{15}\)

\(\Rightarrow\) (2x + 3).15 = (4x - 1).7 (Tính chất tỉ lệ thức)

\(\Rightarrow\) 30x + 45 = 28x - 7

\(\Rightarrow\) 2x = -52

\(\Rightarrow\) x = -26

Chúc bn học tốt!

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2x(X-1)-2x(X+1)=7

=>2x²-2x-2x²-2x=7

=>-4x=7

=>x=-7/4

2x(X-1)- 2x(X+1) = 7

=> 2x(X-1-X+1) =7

<=> 2xX=7

=> x = 7/2 = 3.5

À nhầm, sai rồi :(

Ta có:

8 x 3 - 1 = 2 x 3 - 1 3 = 2 x - 1 . 4 x 2 + 2 x + 1 = - 1 - 2 x . 4 x 2 + 2 x + 1

Do đó, ( 8 x 3 - 1 : 1 - 2 x  = - 4 x 2 + 2 x + 1 = - 4 x 2 - 2 x - 1

Chọn B.  - 4 x 2 - 2 x - 1

Bài làm 

\(\left(2x-1\right)^7=x^7\Leftrightarrow2x-1=x\Leftrightarrow x=1\)

\(\left(2x-1\right)^7=x^7\)
\(2x-1=x\)
2x-x=1

X.(2-1)=1

X.1=1

X=1:1

X=1

Vậy x=1

hi

a) Ta có: \(\left(3x^2-2x+5\right)-\left(x^2+4x^2-x-7\right)\)

\(=3x^2-2x+5-5x^2+x+7\)

\(=-2x^2-x+12\)

b) Ta có: \(4\left(2x+1\right)-5\left(3x+2\right)\)

\(=8x+4-15x-10\)

=-7x-6

hai câu khác nhau,đừng nhầm lẫn

(3x²-2x+5)-(x²+4x²-x-7)

=3x^2 -2x+5-x^2+4x^2-x-7

=6x^2-3x-2