????

Ta có :

\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)

\(\Leftrightarrow\left(x-10\right)^{x+1}\left[1-\left(x-10\right)^{10}\right]=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-10=0\\1-\left(x-10\right)^{10}=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x-10=1\\x-10=-1\end{array}\right.\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x=11\\x=9\end{array}\right.\end{array}\right.\)

Vậy x = 10 ; x = 11 ; x =  9

\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)

\(\Rightarrow\left(x-10\right)^{x+1}.\left[1-\left(x-10\right)^{10}\right]=0\)

\(\Rightarrow\left(x-10\right)^{x+1}=0\) hoặc \(1-\left(x-10\right)^{10}=0\)

+) \(\left(x-10\right)^{x+1}=0\)

\(\Rightarrow x-10=0\)

\(\Rightarrow x=10\)

+) \(1-\left(x-10\right)^{10}=0\)

\(\Rightarrow\left(x-10\right)^{10}=1\)

\(\Rightarrow x-10=\pm1\)

+ \(x-10=1\Rightarrow x=11\)

+ \(x-10=-1\Rightarrow x=9\)

Vậy \(x\in\left\{10;11;9\right\}\)

Bài 2: a: \(\left(x^2-5\right)-\left(x-7\right)\left(x+7\right)\)

\(=x^2-5-\left(x^2-49\right)\)

\(=x^2-5-x^2+49\)

=44

b: \(\left(2x+3y\right)^2+\left(3x-2y\right)^2-2\left(2x+3y\right)\left(3x-2y\right)\)

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

\(=\left(-x+5y\right)^2=x^2-10xy+25y^2\)

Bài 1:

a: \(5x\left(3x^2-4x+2\right)=5x\cdot3x^2-5x\cdot4x+5x\cdot2\)

\(=15x^3-20x^2+10x\)

b: \(\left(x^2-3x\right)\left(3x^2-x+4\right)\)

\(=x^2\cdot3x^2-x^2\cdot x+x^2\cdot4-3x\cdot3x^2+3x\cdot x-3x\cdot4\)

\(=3x^4-x^3+4x^2-9x^3+3x^2-12x=3x^4-10x^3+7x^2-12x\)

Bài 3:

a: \(x^3-6x^2+12x-8=0\)

=>\(\left(x-2\right)^3=0\)

=>x-2=0

=>x=2

b: \(x^3+9x^2+27x+27=0\)

=>\(x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)

=>\(\left(x+3\right)^3=0\)

=>x+3=0

=>x=-3

1. have never gone on a

2. haven't met Alan since

3. remember Mary being

4. brought her up

5. a three-hour drive

6. never arrives late

7. as soon as we

8. grow very slowly

9. she said the truth

10. fed up with her talking

11. did the machine break down

12. was made to

jup mik ik mn ơi huhu

bn viết đề khó hiểu quá

a: Xét ΔAHB vuông tại H và ΔCAB vuông tại A có 

góc B chung

DO đó: ΔAHB\(\sim\)ΔCAB

Suy ra: AH/CA=AB/CB

hay \(AH\cdot BC=AB\cdot AC\)

b: Xét tứ giác ADHE có \(\widehat{ADH}=\widehat{AEH}=\widehat{EAD}=90^0\)

nên ADHE là hình chữ nhật

c: BC=15cm

=>AH=7,2(cm)

mà AH=DE

nên DE=7,2(cm)

a) Xét \(\Delta AHB\) và \(\Delta CAB:\)

\(\widehat{AHB}=\widehat{CAB}\left(=90^o\right).\\ \widehat{ABH}chung.\\ \Rightarrow\Delta AHB\sim\Delta CAB\left(g-g\right).\)

\(\Rightarrow\dfrac{AH}{AC}=\dfrac{AB}{CB}.\\ \Rightarrow AH.CB=AB.AC.\)

b) Xét tứ giác DHEA:

\(\widehat{DAE}=90^o;\widehat{ADH}=90^o;\widehat{AEH}=90^o.\)

\(\Rightarrow\) Tứ giác DHEA là hình chữ nhật.

c) Xét \(\Delta ABC\) vuông tại A:

\(BC^2=AB^2+AC^2\left(Pytago\right).\\ \Rightarrow BC^2=9^2+12^2.\\ \Rightarrow BC=15\left(cm\right).\)

Xét \(\Delta ABC\) vuông tại A; đường cao AH:

\(AH.BC=AB.AC\) (Hệ thức lượng).

\(\Rightarrow AH.15=9.12.\\ \Rightarrow AH=7,2\left(cm\right).\)

Mà \(AH=DE\) (Tứ giác DHEA là hình chữ nhật).

\(\Rightarrow AH=DE=7,2\left(cm\right).\)

\(21_{10}=10101_2\)