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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x=\dfrac{1}{2};y=-100\)
\(\Rightarrow A=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)
\(\Rightarrow A=100\)
\(b,x=-1\)
\(\Rightarrow B=\left[\left(-1\right)^2-5\right]\left(-1+3\right)+\left(-1+4\right)\left[-1-\left(-1\right)^2\right]\)
\(\Rightarrow B=-14\)
\(c,x=-2\)
\(\Rightarrow C=-6\left(5.4-2\right)-5.4\left(7-6\right)-2,5\left(2-14.4\right)\)
\(\Rightarrow C=7\)
\(d,\left|x\right|=2\)
\(TH_1:x\ge0\)
\(D=\left(3.2+5\right)\left(2.2-1\right)+\left(4.2-1\right)\left(3.2+2\right)=89\)
\(TH_2:x< 0\)
\(D=\left(-6+5\right)\left(-4-1\right)+\left(-8-1\right)\left(-6+2\right)=41\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Xét ΔABE vuông tai A và ΔHBE vuông tại H có
BE chung
gócABE=gócHBE
=>ΔABE=ΔHBE
b: ΔBAE=ΔBHE
=>BA=BH; EA=EH
=>BE là trung trực của AH
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(x-2\right)=1\)
\(\Leftrightarrow\left(x^3-3^3\right)-x\left(x+2\right)\left(x-2\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-2^2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left[\left(x+1\right)^2+\left(x^2-1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow6x^2+2-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{12}=-\dfrac{1}{2}\)
\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+\left(x^2+2x\right)\left(2-x\right)-1=0\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+2x^2-x^3+4x-2x^2-1=0\\ \Leftrightarrow x^3-x^3-3x^2+3x^2+2x^2-2x^2-9x+9x+4x=1+27\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: 4x^2+4x+1=(2x+1)^2
c: =(6x+1)^2
d: =(3x-4y)^2
e: =(1/2x+2y)^2
f: =-(x^2-10x+25)
=-(x-5)^2
g: =-(16a^4b^6+24a^5b^5+9a^6b^4)
=-a^4b^4(16b^2+24ab+9a^2)
=-a^4b^2(4b+3a)^2
h: =(5x)^2-2*5x*2y+(2y)^2
=(5x-2y)^2
i: =(5x^2)^2-2*5x^2*y+y^2
=(5x^2-y)^2
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left[\left(3x+1\right)-2\left(x-2\right)\right]\left[\left(3x+1\right)+2\left(x-2\right)\right]\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
c) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)-2\left(x+1\right)\right]\left[3\left(2x+3\right)+2\left(x+1\right)\right]\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x-7\right)\left(8x+11\right)\)
f) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left[2bc-\left(b^2+c^2-a^2\right)\right]\left[2bc+\left(b^2+c^2-a^2\right)\right]\)
\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
g: =(ax+by-ay-bx)(ax+by+ay+bx)
=[a(x-y)-b(x-y)]*[a(x+y)+b(x+y)]
=(x-y)(x+y)(a-b)(a+b)
h: =(a^2+b^2-5-2ab-4)(a^2+b^2-5+2ab+4)
=[(a-b)^2-9][(a+b)^2-1]
=(a-b-3)(a-b+3)(a+b-1)(a+b+1)
i: =(4x^2-3x-18-4x^2-3x)(4x^2-3x-18+4x^2+3x)
=(-6x-18)(8x^2-18)
=-12(x+3)(4x^2-9)
=-12(x+3)(2x-3)(2x+3)
k: =(3x+3y-3)^2-(4x+6y+2)^2
=(3x+3y-3-4x-6y-2)(3x+3y-3+4x+6y+2)
=(-x-3y-5)(7x+9y-1)
i: =25-(2x-3y)^2
=(5-2x+3y)(5+2x-3y)
m: =(x-y)^2-(2m-n)^2
=(x-y-2m+n)(x-y+2m-n)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3\left(2a-1\right)+5\left(3-a\right)\)
\(=6a-3+15-5a\)
\(=a-12\)
Thay \(a=\dfrac{-3}{2}\) vào biểu thức trên
\(a-12\)
\(=\dfrac{-3}{2}-12\)
\(=\dfrac{-27}{2}\)
\(b,25x-4\left(3x-1\right)+7\left(5-2x\right)\)
\(=25x-12x+4+35-14x\)
\(=-1x+39\)
Thay \(x=2,1\) vào biểu thức trên
\(-1x+39\)
\(=-1.2,1+39\)
\(=-2,1+39\)
\(=36,9\)
\(c,4a-2\left(10a-1\right)+8a-2\)
\(=4a-20a+2+8a-2\)
\(=-8a\)
Thay \(a=-0,2\) vào biểu thức trên
\(-8a\)
\(=-8.\left(-0,2\right)\)
\(=1,6\)
\(d,12\left(2-3b\right)+35b-9\left(b+1\right)\)
\(=24-36b+35b-9b-9\)
\(=-10b-15\)
Thay \(b=\dfrac{1}{2}\) vào biểu thức trên
\(-10b-15\)
\(=-10.\dfrac{1}{2}-15\)
\(=-20\)
`9,`
`a, 5/6+1/6 \div 4/3`
`= 5/6+1/8`
`= 23/24`
`b,`
`15/4*(-2/3+3/4)+15/4*(-1/3+1/4)`
`= 15/4*[(-2/3+3/4)+(-1/3+1/4)]`
`= 15/4*[(-2/3)+3/4-1/3+1/4]`
`= 15/4*[(-2/3-1/3)+(3/4+1/4)]`
`= 15/4*(-1+1)`
`= 15/4*0=0`
`10,`
`a, 35 - 3(x-30)=20`
`35- 3x-90=20`
`35+90-3x=20`
`125-3x=20`
`3x=125-20`
`3x=105`
`x=105 \div 3`
`x=35`
`b,`
`|2x-1/2|=3/2`
`=>`\(\left[{}\begin{matrix}2x-\dfrac{1}{2}=\dfrac{3}{2}\\2x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=\dfrac{3}{2}+\dfrac{1}{2}\\2x=-\dfrac{3}{2}+\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=2\\2x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)