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\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
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= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
\(\left(4A\right)\\ a,\\ \Leftrightarrow\left[\left(x-2\right)\left(2x+3\right)\right]\left[\left(x-2\right)\left(2x+3\right)\right]=0\\ \Leftrightarrow\left(-x-5\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x-5=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{-1}{3}\end{matrix}\right.\\ b,\\ \Leftrightarrow\left[3\left(2x+1\right)\right]^2-\left[2\left(x+1\right)\right]^2=0\\ \Leftrightarrow\left[3\left(2x+1\right)-2\left(x+1\right)\right]\left[3\left(2x+1\right)+2\left(x+1\right)\right]=0\\ \Leftrightarrow\left(4x+1\right)\left(8x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-5}{8}\end{matrix}\right.\\ c,\\ \Leftrightarrow\left[\left(x+1\right)+1\right]^2=0\\ \Leftrightarrow\left(x+1\right)+1=0\\ \Leftrightarrow x+2=0\Rightarrow x=-2\\ d,\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)+\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left[\left(x-1\right)\left(x+3\right)+1\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(\left(4B\right)\\ a,\\ \Leftrightarrow49-14x+x^2-4\left(x+25\right)^2=0\\ \Leftrightarrow49-14x+x^2-4x^2-40x-100=0\\ \Leftrightarrow3x^2-54x-51=0\\ \Leftrightarrow-3\left(x^2+18x+17\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+17\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-17\end{matrix}\right.\\ b,\\ \Leftrightarrow4x^2\left(x^2-2x+1\right)-\left(4x^2+4x+1\right)=0\\ \Leftrightarrow x^2-6x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
\(c,\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(2-x\right)=0\\ \Leftrightarrow\left(x+1\right)\left[\left(x^2-x+1\right)-\left(2-x\right)\right]=0\\ \Leftrightarrow\left(x+1\right)\left(x^1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-1\end{matrix}\right.\\ d,\\ \Leftrightarrow\left(x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
\(x^4-10x^3+35x^2+24>0\)
\(\Leftrightarrow x^4-2.5.x^3+\left(5x\right)^2+10x^2+24>0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+10x^2+24>0\)
\(\Leftrightarrow x^2\left(x-5\right)^2+10x^2+24>0\)(luôn đúng)
Vậy nghiệm của bất phương trình \(x\in R\)
\(10x-25-x^2=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.x.5+5^2\right)=-\left(x-5\right)^2\)
5: \(=4b^2-2b+\dfrac{1}{4}-\dfrac{1}{4}+a-a^2\)
\(=\left(2b\right)^2-2\cdot2b\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(a^2-a+\dfrac{1}{4}\right)\)
\(=\left(2b-\dfrac{1}{2}\right)^2-\left(a-\dfrac{1}{2}\right)^2\)
\(=\left(2b-\dfrac{1}{2}-a+\dfrac{1}{2}\right)\left(2b-\dfrac{1}{2}+a-\dfrac{1}{2}\right)\)
\(=\left(2b-a\right)\left(2b+a-1\right)\)
6:
\(=b^2-4b+4-9c^2\)
\(=\left(b-2\right)^2-9c^2\)
\(=\left(b-2-3c\right)\left(b-2+3c\right)\)
\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right).\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)
\(x^2-10x-11=0\)
=>\(x^2-10x+25-36=0\)
=>\(\left(x-5\right)^2-6^2=0\)
=>(x-5-6)(x-5+6)=0
=>(x-11)(x+1)=0
=>\(\left[{}\begin{matrix}x-11=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
\(x^2\)\(-2.x.5+5^2\)\(-36\)\(=0\)
\(\Leftrightarrow\)\(\left(x-5\right)^2\)\(-36=0\)
\(\Leftrightarrow\left(x-5^{ }\right)^2\)\(=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)