Làm cho bạn 1 con thôi dài quá trôi hết màn hình:

c) có vẻ khó nhất (con khác tương tự)

đặt 2x+2=t=> x+1=t/2

\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)

\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)

<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)

Giải phương trình đó mọi người

len google di ban

mk chua hoc bai nay

d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)

\(=8x^2+12x-8x^2-32x\)

=-20x

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

\(=10x^2+4x+6x^2-2x-9x+3\)

\(=16x^2-7x+3\)

f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)

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

Câu a:

\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

\(\Leftrightarrow\left(64x^2-16x+1\right)\left(64x^2-16x\right)=72\)

Đặt 64x² - 16x = t \(\left(t\ge-1\right)\)

\(\Rightarrow t\left(t+1\right)=72\)

\(\Leftrightarrow\left(t+9\right)\left(t-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-9\left(loai\right)\\t=8\left(nhan\right)\end{matrix}\right.\)

\(\Rightarrow64x^2-16x=8\)

\(\Leftrightarrow8\left(2x-1\right)\left(4x+1\right)=0\)

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

Câu b:

\(\Leftrightarrow\left(x+1\right)^2\left(2x+1\right)\left(2x+3\right)=18\)

\(\Leftrightarrow\left(4x^2+8x+4\right)\left(4x^2+8x+3\right)=72\)

Đặt 4x² + 8x + 4 = m \(\left(m\ge0\right)\)

\(\Rightarrow m\left(m-1\right)=72\)

\(\Leftrightarrow\left(m-9\right)\left(m+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=9\left(nhan\right)\\m=-8\left(loai\right)\end{matrix}\right.\)

\(\Rightarrow4\left(x+1\right)^2=9\)

\(\Leftrightarrow x+1=\pm\dfrac{3}{2}\)

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

1. (3x - 5)² - (3x + 1)² = 8

=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8

=> -6(6x - 4) = 8

=> 6x - 4 = \(\dfrac{-4}{3}\)

\(\Rightarrow x=\dfrac{4}{9}\)

2) 2x(8x - 3) - (4x - 3)² = 27

=> 16x² - 6x - 16x² + 24x - 9 = 27

=> 18x - 9 = 27

=> x = 2

3) (2x - 3)² - (2x + 1)² = 3

=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3

=> -4(4x - 2) = 3

=> 4x - 2 = \(\dfrac{-3}{4}\)

\(\Rightarrow x=\dfrac{5}{16}\)

4) (x + 5)² - x² = 45

=> (x + 5 - x)(x + 5 + x) = 45

=> 5(2x + 5) = 45

=> 2x + 5 = 9

=> x = 2

5) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 18

=> x³ - 9x² + 27x - 27 - x³ + 27 + 9(x² + 2x + 1) = 18

=> -9x² + 27x + 9x² + 18x + 9 = 18

=> 45x + 9 = 18

=> 45x = 9

=> x = \(\dfrac{1}{5}\)

6) x(x - 4)(x + 4) - (x - 5)(x² + 5x + 25) = 13

=> x (x² - 16) - (x³ - 125) = 13

=> x³ - 16x - x³ + 125 = 13

=> -16x = -112

=> x = 7.