\(3x^4-4x^3+2x\left(x^3-2x^2+7x\right)\)

\(=3x^4-4x^3+2x^4-4x^3+14x^2\)

\(=5x^4-8x^3+14x^2\)

3x⁴ - 4x³ + 2x(x³ - 2x² + 7x )

= 3x⁴ - 4x³ + 2x⁴ _ 4x³ + 14x²

= 5x⁴ - 8x³ + 14x²

cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?

a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ( x + 4 )² - ( x + 1 )( x - 1 ) = 16

<=> x² + 8x + 16 - ( x² - 1 ) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = -1/8

b) ( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 49 ) = 0

<=> 5x² + 2x + 10 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> 2x = -255/2

c) ( x + 2 )( x² - 2x + 4 ) - x( x² + 2 ) = 15

<=> x³ + 2³ - x³ - 2x = 15

<=> 8 - 2x = 15

<=> 2x = -7

<=> x = -7/2

d) ( x + 3 )³ - x( 3x + 1 )² + ( 2x + 1 )( 4x² - 2x + 1 ) = 28

<=> x³ + 9x² + 27x + 27 - x( 9x² + 6x + 1 ) + [ ( 2x )³ + 1³ ] = 28

<=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

<=> 3x² + 26x + 28 = 28

<=> 3x² + 26x = 0

<=> x( 3x + 26 ) = 0

<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)

a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)

b/

\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)

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

\(\Leftrightarrow\left|x+2\right|-4=0\)

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

c/

\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)

\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: ...

\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)

Đặt \(\frac{\left|x-2\right|}{x-1}=a\)

\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)

e/ ĐKXĐ: ...

Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)

\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)

f/ ĐKXĐ: ...

Đặt \(\left|x-\frac{1}{x}\right|=a\ge0\Rightarrow a^2=x^2+\frac{1}{x^2}-2\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

Phương trình trở thành:

\(a^2+2-10=2a\)

\(\Leftrightarrow a^2-2a-8=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|x-\frac{1}{x}\right|=4\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-1=0\\x^2+4x-1=0\end{matrix}\right.\)