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a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
=\(\frac{1}{x+1}-\frac{1}{x+2}\)+\(\frac{1}{x+2}-\frac{1}{x+3}\)+\(\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}\)-\(\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
=\(\frac{1}{x+1}-\frac{1}{x+6}\)
=\(\frac{x+6-x-1}{\left(x+1\right)\left(x+6\right)}\)
=\(\frac{5}{\left(x+1\right)\left(x+6\right)}\)
\(A=x^3\left(x^3-3x^2+3x-1\right)+\left(x^4-2x^3+x^2\right)+\left(x^2-x\right)+1\)
\(=x^3\left(x-1\right)^3+\left(x^2-x\right)^2+\left(x^2-x\right)+1\)
\(=\left(x^2-x\right)^3+\left(x^2-x\right)^2+\left(x^2-x\right)+1\)
\(=10^3+10^2+10+1=1111\)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
a, \(E=\left(\frac{x^2+4}{x^2-4}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)
\(=\left(\frac{x^2+4}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2+4-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{6}{x+2}\right)\)
\(=\frac{x^2+4-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{x^2-x-2}{6\left(x-2\right)}=\frac{x+1}{6}\)
b, Ta có : \(\left|2x-3\right|=1\Leftrightarrow\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(ktmđk\right)\\x=1\end{cases}}}\)
Thay x = 1 vào biểu thức E ta được : \(\frac{1+1}{6}=\frac{2}{6}=\frac{1}{3}\)
Vậy với x = 1 thì E = 1/3
c, Ta có : \(E< 0\)hay \(\frac{x+1}{6}< 0\Rightarrow x+1>0\)( do 6 > 0 )
\(\Leftrightarrow x>-1\)
Với với x > -1 thì E < 0
d, Ta có E = 3 - x hay \(\frac{x+1}{6}=3-x\Rightarrow x+1=18-6x\Leftrightarrow7x=17\Leftrightarrow x=\frac{17}{7}\)
a: \(=\dfrac{6}{x+1}+\dfrac{4}{x-1}-\dfrac{10}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x-6+4x+4-10}{\left(x-1\right)\left(x+1\right)}=\dfrac{10x-12}{\left(x-1\right)\left(x+1\right)}\)
b: \(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{6}{x}=\dfrac{1}{x}+\dfrac{6}{x}=\dfrac{7}{x}\)