a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

Bạn nhân đa thức với đa thức

Theo bài ra, ta suy ra được:

32x^5 +1 -(32x^5 -1) =2

2 = 2

Vậy có vô số x thỏa mãn đề bài.

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

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

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

\(4+2x.2x+8x=-x\)

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

\(13x=-4\)

\(x=-\frac{4}{13}\)

 Vậy pt có nghiệm là { -4/13 }

2) mình nghĩ thế này

(2x-3)^2=2x-3

Đẻ 2 cái trên = nhau thfi 

2x-3=1

=> x=2

 4-3=2( dân chơi mới hiểu)

Chắc là viết thiếu số "1" đấy, sợ lớp 11 còn chưa làm được cơ