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a,
\(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
b,
\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)
c,
a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right).x^2.\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
Ta có:
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)x^2\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left[\left(2x^3-x^2+1\right)+\left(x^2-2x^3\right)\right]\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
1) \(\left(x+2\right)^3-\left(x+6\right)^2-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+3.x^2.2+3.x.2^2+2^3-\left(x^2+2.x.6+6^2\right)-\left(x^3+1\right)\)
\(=x^3+6x^2+12x+8-x^2-12x-36-x^3-1\)
\(=5x^2-29\)
2. \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+3\right)^2\)
\(=x^3-3^3-\left(x^2+2.x.3+3^2\right)\)
\(=x^3-27-x^2-6x-9\)
\(=x^3-x^2-6x-36\)
\(=x^3-2x^2+3x^2-6x-36\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)-36\)
\(=x\left(x-2\right)\left(x+3\right)-36\)
3. \(\left(2x-1\right)^2+2x^2.\left(x-2\right)\left(x+3\right)^2\)
\(=4x^2-4x+1+2x^2\left(x-2\right)\left(x^2+6x+9\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+6x^2+9x-2x^2-12x-18\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+4x^2-3x-18\right)\)
\(=4x^2-4x+1+2x^5+8x^4-6x^3-36x^2\)
\(=2x^5+8x^4-6x^3-32x^2-4x+1\)
....
P/s: Không chắc lắm
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