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

\(g,\left(x^2y-xy+xy^2+y^3\right).3xy^2\\ =\left(3xy^2.x^2y\right)-\left(3xy^2.xy\right)+\left(3xy^2.xy^2\right)+\left(3xy^2.y^3\right)\\ =3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\)

\(h,\dfrac{2}{3}x^2y\left(15x-0,9y+6\right)\\ =\left(\dfrac{2}{3}x^2y.15x\right)-\left(\dfrac{2}{3}x^2y.0,9y\right)+\left(\dfrac{2}{3}x^2y.6\right)\\ =10x^3y-\dfrac{3}{5}x^2y^2+4x^2y\)

\(i,-\dfrac{3}{7}x^4\left(2,1y^2-0,7x+35\right)\\ =\left(-\dfrac{3}{7}x^4.2,1y^2\right)-\left(-\dfrac{3}{7}x^4.0,7x\right)+\left(-\dfrac{3}{7}x^4.35\right)\\ =-\dfrac{9}{10}x^4y^2+\dfrac{3}{10}x^5-15x^4\)

g: =x^2y*3xy^2-xy*3xy^2+xy^2*3xy^2+y^3*3xy^2

=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5

h: =2/3x^2y*15x-2/3x^2y*0,9y+2/3x^2y*6

=10x^3y-0,6x^2y^2+4x^2y

i: =-3/7x^4*2,1y^2+3/7x^4*0,7x-3/7x^4*35

=-0,9x^4y^2+3/10x^5-15x^4

\(a.x^3+12x^2+48x+64=x^3+3.4x^2+3.4^2x+4^3=\left(x+4\right)^3\)

Thay \(x=6\) vào \(\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

\(b,B=x^3-6x^2+12x-8=\left(x-2\right)^3\)

Thay \(x=22\) vào \(\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

\(c,C=x^3+9x^2+27x+27=x^3+3.3x^2+3.3^2x+3^3=\left(x+3\right)^3\)

Thay \(x=-103\) vào \(\left(x+3\right)^3=\left(-103+3\right)^3=\left(-100\right)^3=-1000000\)

\(d,D=x^3-15x^2+75x-125=x^3-3.5x^2+5^2.3x-5^3=(x-5)^3\)

Thay \(x=25\) vào \(\left(x-5\right)^3=\left(25-5\right)^3=20^3=8000\)

a) \(A=x^3+12x^2+48x+64\)

\(=x^3+3\cdot4\cdot x^2+3\cdot4^2\cdot x+4^3\)

\(=\left(x+4\right)^3\)

Thay \(x=6\) vào biểu thức A ta có:

\(\left(6+4\right)^3=10^3=1000\)

Vậy: ...

b) \(B=x^3-6x^2+12x-8\)

\(=x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3\)

\(=\left(x-2\right)^3\)

Thay \(x=22\) vào biểu thức B ta có:

\(\left(22-2\right)^3=20^3=8000\)

Vậy: ...

c) \(C=x^3+9x^2+27x+27\)

\(=x^3+3\cdot3\cdot x^2+3\cdot3^2\cdot x+3^3\)

\(=\left(x+3\right)^3\)

Thay \(x=-103\) vào biểu thức C ta được:

\(\left(-103+3\right)^3=\left(-100\right)^3=-1000000\)

Vậy: ...

d) \(D=x^3-15x^2+75x-125\)

\(=x^3-3\cdot5\cdot x^2+3\cdot5^2\cdot x-5^3\)

\(=\left(x-5\right)^3\)

Thay \(x=25\) vào biểu thức D ta được:

\(\left(25-5\right)^3=20^3=8000\)

Vậy: ...

\(a,x=\dfrac{1}{2};y=-100\)

\(\Rightarrow A=\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+100\right]-\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-100\right)-100\left[\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

\(\Rightarrow A=100\)

\(b,x=-1\)

\(\Rightarrow B=\left[\left(-1\right)^2-5\right]\left(-1+3\right)+\left(-1+4\right)\left[-1-\left(-1\right)^2\right]\)

\(\Rightarrow B=-14\)

\(c,x=-2\)

\(\Rightarrow C=-6\left(5.4-2\right)-5.4\left(7-6\right)-2,5\left(2-14.4\right)\)

\(\Rightarrow C=7\)

\(d,\left|x\right|=2\)

\(TH_1:x\ge0\)

\(D=\left(3.2+5\right)\left(2.2-1\right)+\left(4.2-1\right)\left(3.2+2\right)=89\)

\(TH_2:x< 0\)

\(D=\left(-6+5\right)\left(-4-1\right)+\left(-8-1\right)\left(-6+2\right)=41\)

B nhé

`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`

`=>`\(\left[{}\begin{matrix}2x-\dfrac{1}{2}=\dfrac{3}{2}\\2x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=\dfrac{3}{2}+\dfrac{1}{2}\\2x=-\dfrac{3}{2}+\dfrac{1}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=2\\2x=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

e: =>(3x-1-x-5)(3x-1+x+5)=0

=>(2x-6)(4x+4)=0

=>x=3 hoặc x=-1

f: =>(2x-1-x+3)(2x-1+x-3)=0

=>(x+2)(3x-4)=0

=>x=4/3 hoặc x=-2

g: =>4x^2-4x+1-4x^2+1=0

=>-4x+2=0

=>x=1/2

g: =>(x^2+4)(x^2-1)=0

=>x^2-1=0

=>x=1 hoặc x=-1

m: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

l: =>(x-2)(x^2+2x+4-x+12)=0

=>(x-2)(x^2+x+16)=0

=>x-2=0

=>x=2

k: =>(2x-5)(2x+5-2x-7)=0

=>2x-5=0

=>x=5/2

\(a.x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

\(b.3x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\end{matrix}\right.\)

\(c.x\left(x-4\right)-2x+8=0\)

\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(d.3x\left(x+5\right)-3x-15=0\)

\(\Leftrightarrow3x\left(x+5\right)-3\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(d'.x\left(2x-3\right)-3\left(3-2x\right)=0\)

\(\Leftrightarrow x\left(2x-3\right)+3\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)

a: =>(x+5)(x-5-1)=0

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: =>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

c: =>(x-4)(x-2)=0

=>x=4 hoặc x=2

d: =>(x+5)(3x-5)=0

=>x=5/3 hoặc x=-5

e: =>x(2x-3)+3(2x-3)=0

=>(2x-3)(x+3)=0

=>x=3/2 hoặc x=-3

34 x 54 = 1836 nha

Nhớ k nhé

34x54=1836

HT