Bài 1: 

a) Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(AH^2=BH\cdot CH\)

\(\Leftrightarrow x=\dfrac{6^2}{12}=\dfrac{36}{12}=3\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABH vuông tại H, ta được:

\(AB^2=AH^2+BH^2\)

\(\Leftrightarrow y^2=3^2+6^2=9+36=45\)

hay \(y=3\sqrt{5}\left(cm\right)\)

b) Ta có: \(\sin^2\widehat{A}+\cos^2\widehat{A}=1\)

\(\Leftrightarrow\cos^2\widehat{A}=\dfrac{3}{4}\)

hay \(\cos\widehat{A}=\dfrac{\sqrt{3}}{2}\)

Ta có: \(\tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}\)

\(=\dfrac{1}{2}:\dfrac{\sqrt{3}}{2}\)

\(=\dfrac{\sqrt{3}}{3}\)

Lời giải:

a. Theo định lý Pitago cho tam giác $ABH$:

$y^2=36+x^2(1)$

Theo hệ thức lượng trong tam giác vuông:

$y^2=x(x+12)(2)$

Từ $(1);(2)\Rightarrow x^2+36=x(x+12)$

$36=12x$

$x=3$ (cm)

$y^2=3(3+12)=45\Rightarrow y=3\sqrt{5}=6,708$ (cm)

b.

Giả sử $a = \widehat{B}$

$\cos a=\cos B=\frac{AB}{BC}=\frac{1}{2}$

$\Rightarrow AB=\frac{BC}{2}$

Theo Pitago: $AC=\sqrt{BC^2-AB^2}=\sqrt{BC^2-(\frac{BC}{2})^2}=\frac{\sqrt{3}}{2}BC$

$\sin a= \sin B=\frac{AC}{BC}=\frac{\sqrt{3}}{2}$

$\tan a=\tan B=\frac{AC}{AB}=\frac{\frac{BC}{2}}{\frac{\sqrt{3}}{2}BC}=\frac{\sqrt{3}}{3}$

Bài 2: 

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3

Vậy: A ⋮ 3

_____________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7

Vậy: A ⋮ 7

___________________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5

Vậy: A ⋮ 5 

mn giúp e với

\(\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

bài 2

a)

\(2xy^2-4y\\ =2y\left(xy-2\right)\)

b)

\(x^2-6xy+9y^2\\ =\left(x-3y\right)^2\)

c)

\(x^2+x-y^2+y\\ =\left(x^2-y^2\right)+\left(x+y\right)\\ =\left(x-y\right)\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(x-y+1\right)\)

d)

\(x^2+4x+3\\ =x^2+3x+x+3\\ =x\left(x+3\right)+\left(x+3\right)\\ =\left(x+3\right)\left(x+1\right)\)

4,\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)

2,\(=y\left(x^2-6x+9\right)=y\left(x-3\right)^2\)

1,=\(2y\left(xy-2\right)\)

Ta có: \(\widehat{B_1}=\widehat{B_2}=50^0\)(đối đỉnh)

\(\Rightarrow\widehat{A_1}=\widehat{B_2}=50^0\)

Mà 2 góc này so le trong

=> a//b

Ta có: A1 = B1 = 50 độ

Mà 2 góc này ở vị trí đồng vị

=> a//b

a. =[(-4)(-25)].3=100.3=300

b. =[(-8).125][5.2]=-1000.10=-10000

c. =[25.(-4)][(-5).20]=-100(-100)=10000

bủh

a: \(\sqrt{5a^2}=\left|a\sqrt{5}\right|=-a\sqrt{5}\left(a< =0\right)\)

c: A=\(\sqrt{72a^2b^4}=\sqrt{36a^2b^4\cdot2}=6\sqrt{2}\cdot b^2\cdot\left|a\right|\)

mà a<0

nên \(A=-6\sqrt{2}\cdot ab^2\)

d: \(\sqrt{24a^4b^8}=\sqrt{4a^4b^8\cdot6}=2a^2b^4\cdot\sqrt{6}\)

B1:

\(B\left(7\right)=\left\{0;\pm7;\pm14;....\right\}\\ B\left(-7\right)=\left\{0;\pm7;\pm14;....\right\}\\ B\left(5\right)=\left\{0;\pm5;\pm10;...\right\}\\ B\left(-8\right)=\left\{0;\pm8;\pm16;...\right\}\)

B2:

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\\ Ư\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\ Ư\left(36\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm9;\pm12;\pm18;\pm36\right\};Ư\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)