A B C H

Xét tam giác ABH ta có : \(AB^2=BH^2+AH^2\Rightarrow15^2=12^2+AH^2\)

\(AH^2=AB^2-BH^2=15^2-12^2=81\Rightarrow AH=9\)cm 

Xét tam giác ACH ta có : \(AC^2=AH^2+HC^2\Rightarrow AC^2=AH^2+HC^2\)

\(HC^2=AC^2-AH^2=41^2-9^2=1600\Rightarrow HC=40\)cm 

Ta có : \(BC=CH+HB=40+12=52\)cm

\(\Rightarrow S_{ABC}=\frac{1}{2}AH.BC=\frac{1}{2}.9.52=234\)cm²

a, \(x^2-\frac{x^4}{x^2}-1=x^2-x^2-1=-1\)

b, \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x+3\right)}\)

\(=\frac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\frac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x\left(x-3\right)}\)

c, \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x\left(1-2x\right)}\)

\(=\frac{1-3x}{2x}-\frac{3x-2}{1-2x}+\frac{3x-2}{2x\left(1-2x\right)}=\frac{\left(1-3x\right)\left(1-2x\right)}{2x\left(1-2x\right)}-\frac{2x\left(3x-2\right)}{2x\left(1-2x\right)}+\frac{3x-2}{2x\left(1-2x\right)}\)

\(=\frac{1-2x-3x+6x^2-6x^2+4x+3x-2}{2x\left(1-2x\right)}=\frac{-1+2x}{2x\left(1-2x\right)}=\frac{-\left(1-2x\right)}{2x\left(1-2x\right)}=\frac{-1}{2x}\)

d, viết lại đề đy nhé 

e, \(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}=\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x-2x^2}{\left(3-x\right)\left(x+3\right)}\)

\(=\frac{x+1}{x-3}-\frac{1-x}{x+3}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+3x+x+3-x+3+x^2+3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{8x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\left(a-\sqrt{2011}\right)\left(b+\sqrt{2011}\right)=14\)

\(\Leftrightarrow ab+\sqrt{2011}\left(a-b\right)=2025\)

Có: a,b nguyên => a-b nguyên

=> VP=VT <=> \(\sqrt{2011}\left(a-b\right)\)nguyên

=> a-b=0 <=> a=b

=> pt <=> a^2=2025

Làm nốt nhé. 

k ko biết

treen toán ko dc đưa những hình ảnh này. OK

+= (-45+45) + ( -13+-17+20)+48

= 0=0+48=48

+ 100-50/25

=100-2

=98

+= (-3+3)+( -15+20)

=0+-5

=5

+  2x=73-(-17)

2x=90

x=90/2

x=45

tặng cho mik mấy cái tk nha thanks 

45+(-13)+(-17)+(-45)+20+48=38

100-50/25=98

tìm x

2x+(-17)=73

x=45