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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 \(x=\dfrac{3}{2}\)
a, (1-x)(5x+3)= (3x-8)(1-x)
<=> (1-x) (5x+3) - (3x-8)(1-x) =0 <=> (1-x) (2x+11) = 0
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy.........
b, (x-3)(x+4)-2(3x-2)=(x-4)^2
<=> 3x = 24<=> x=8
Vậy .......
c,x^2+ x^3+x+1=0
<=> x^2 (x+1) +(x+1) =0 <=> (x^2 +1)(x+1) =0
<=> x+1 =0 => x=-1
Vậy.......
d, \(\dfrac{x-3}{x+3}-\dfrac{2}{x-3}=\dfrac{3x+1}{9-x^2}\)
\(\Leftrightarrow x^2-6x+9-2x-6=-3x-1\)
\(\Leftrightarrow x^2-5x+4=0\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
Vậy...........
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
Thank You !