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

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

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

a: \(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CB}\right|=10a\)

b: \(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\dfrac{BC}{2}=5a\)

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

coming 

ỉon        doing 

wakes

buying

có [x-y]2=1

suy ra [x-y]mũ 2= 1 mũ 2

suy ra x-1=1

x=1+1

x=2

x = 2 nha bạn

9C

7D

6D

8D

11A

10C

13B

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)

Câu 1: Vì (d') vuông góc với (d) nên \(a\cdot\dfrac{-1}{3}=-1\)

hay a=3

Vậy: (d'): y=3x+b

Thay x=4 và y=-5 vào (d'), ta được:

b+12=-5

hay b=-17