a) Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)

\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b,\(< =>25x^2+10x+1-25x^2+9-30=0\)

\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)

c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)

\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)

\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)

\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)

a: Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

\(\left(x^2-5x\right)+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

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

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

Bài 10:

a) (x+2)² -x(x+3) + 5x = -20

=> x² + 4x + 4 - x² - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x³-10x²+5x=0   

=>5x(x²-2x+1)=0

=>5x(x-1)² =0

=> 5x=0 hoặc (x-1)²=0

=>x=0 hoặc x=1

c) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

=> (x² - 1)[(x² - 1)² -  (x⁴ + x² + 1)] = 0

<=> (x² - 1)(x⁴ - 2x² + 1 - x⁴ - x² - 1) = 0

<=>  (x² - 1)(-3x²) = 0

<=> (x² - 1)=0 hoặc (-3x²) =0

<=> x²=1 hoặc x²=0

<=> x=−1;1 hoặc x=0

d)

⇔x=-23/12

Học tốt nhé:333

\(a.\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4x-4=5\)

\(\left(-4x-6x\right)+\left(4-9\right)-4x-4=5\)

\(-10x-5-4x-4=5\)

\(-14x-9=5\)

\(-14x=14\Rightarrow x=-1\)

\(b.\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(17x-10=-44\)

\(17x=-34\Rightarrow x=-2\)

\(c.\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(10x+10=30\)

\(10x=20\Rightarrow x=2\)

\(d.\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)

\(\left(x^2+6x+9\right)+\left(x^2-4\right)-2\left(x^2-2x+1\right)=7\)

\(2x^2+6x+5-2x^2+4x-2=7\)

\(10x+3=7\)

\(10x=4\Rightarrow x=\frac{4}{10}=\frac25\)

\(f.\left(3x-8\right)^2=0\)

\(3x-8=0\Rightarrow x=\frac83\)

\(e.6\left(x+1\right)^2-2\left(x+1\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)

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

\(6x^2+12x+6-2x-2+2x^3-2=0\)

\(2x^3+6x^2+10x+2=0\)

\(\Rightarrow x\approx-0,23\)

• 2(x+5)(x-5)-(x+2)(2x-3)+x(x^2-8)=(x+1)(x^2-x+1)

<=> 2(x^2-25) - 2x^2+3x-4x+6 + x^3-8x = x^3+1

=>2x^2-50 - 2x^2 -9x+6+x^3-x^3-1 = 0

<=>-9x - 45 =0

<=>-9x=45

<=>x=-5

Còn phần b và c bạn cứ khai triển ra,mình phải đi học nên không có thời gian giải cho bạn