Bài 1:ĐKXĐ: x>0; x<>1; x<>1/4

a: Ta có: \(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)

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

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

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

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

TA có: \(A=\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\cdot\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\frac{2\sqrt{x}-1}{-x+\sqrt{x}}\)

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

b: \(A=\frac23\)

=>\(\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac23\)

=>\(2x-2\sqrt{x}+2=3\sqrt{x}\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(2x-4\sqrt{x}-\sqrt{x}+2=0\)

=>\(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

=>x=4(nhận) hoặc x=1/4(loại)

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

\(y=\sqrt{6^2+12^2}=6\sqrt{5}\)

b: \(x=\sqrt{4\cdot9}=6\)

c: \(x=5\cdot\tan40^0\simeq4,2\left(cm\right)\)

ghi đầy đủ đc ko ạ

a: Tỉ số là 3/2

b: Tỉ số phần trăm là;

40/(30+40+20+20+5)=34,78%

Câu 1:
TXĐ:D=R

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

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

=>f(x) là hàm số chẵn

mik chưa học đến bài đó nhưng chúc b mai thi tốt nhé!

\(a,ĐK:x\ne0;x\ne5\\ B=\dfrac{x^2-25+2x^2-12x-x^2+8x+25}{2x\left(x-5\right)}=\dfrac{2x\left(x-2\right)}{2x\left(x-5\right)}=\dfrac{x-2}{x-5}\\ b,x=3\Leftrightarrow A=\dfrac{3+6}{5-3}=\dfrac{9}{2}\\ c,\text{Câu a}\\ d,E=B-A=\dfrac{x-2}{x-5}+\dfrac{x+6}{x-5}=\dfrac{2x+4}{x-5}=\dfrac{2\left(x-5\right)+14}{x-5}=2+\dfrac{14}{x-5}\in Z\\ \Leftrightarrow x-5\inƯ\left(14\right)=\left\{-14;-7;-2;-1;1;2;7;14\right\}\\ \Leftrightarrow x\in\left\{-9;-2;3;4;6;7;12;19\right\}\)

Câu 1: 

TXĐ: D=R

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

Vậy: f(x) là hàm số chẵn

Mình cảm ơn ạ

Ai giúp mình với đi ạ 

exs1

1 don't have to

3 mustn't

7  don't have to

exs2

3 might

4 can't

7 might be

còn nữa ...

17. coming 18. iron___ doing 19. wakes 20. buying

coming 

ỉon        doing 

wakes

buying