Giải phương trình sau

1) x³+1=2\(\sqrt[3]{2x-1}\)

2) 2(x²+2)=5\(\sqrt{x^3+1}\)

3) \(\sqrt[3]{\left(3x+1\right)^2}\)+\(\sqrt[3]{\left(3x-1\right)^2}\)+\(\sqrt[3]{9x^2-1}\)=1

4)2( x²-3x+2) =3\(\sqrt{x^3+8}\)

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

6) \(\sqrt{x-1}\)+\(\sqrt{x^3+x^2+x+1}\)=1+\(\sqrt{x^4-1}\)

7) (2x+7)\(\sqrt{2x+7}\)=x²+9x+7

8) \(\sqrt[3]{x+10}\)+\(\sqrt[3]{17-x}\)=3

bài 1: giải các hệ phương trình

1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\)

x+y=9

2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\)

\(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\)

3)\(2|x|-y=3\)

\(|x|+y=3\)

4)\(2\left(x+y\right)+\sqrt{x+1}=4\)

\(\left(x+y\right)-3\sqrt{x+1}=-5\)

5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\)

\(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\)

6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\)

7) \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

\(\dfrac{3}{x}-\dfrac{1}{y}=2\)

8)\(\dfrac{1}{x+2}+\dfrac{3}{2y-1}=4\)

\(\dfrac{4}{x+2}-\dfrac{1}{2y-1}=3\)

9)\(\dfrac{4}{x+y} +\dfrac{1}{y-1}=5\)

\(\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\)

10)\(\dfrac{7}{\sqrt{2x+3}}-\dfrac{4}{\sqrt{3}-y}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{2x+3}}+\dfrac{3}{\sqrt{3-y}}=\dfrac{13}{6}\)

11)\(\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\)

\(\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\)

12) \(\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}2\dfrac{1}{6}\)

13) \(3\sqrt{x-1}+2\sqrt{y}=13\)

\(2\sqrt{x-1}-\sqrt{y}=4\)

14) 6x + 6y = 5xy

\(\dfrac{4}{x}-\dfrac{3}{y}=1\)

Làm mất căn mẫu và thu gọn

1) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

2) \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

3) \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\dfrac{1}{\sqrt{5}}\)

4) \(\dfrac{1}{1+\sqrt{2}}-\dfrac{1}{1-\sqrt{2}}\)

5) \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{\sqrt{5}-3}\)

6) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\)

7) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)

8) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3}{\sqrt{2}-1}\)

9) \(\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}-1}-\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}+1}\)

10) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)

11) \(\dfrac{5}{1+\sqrt{6}}-\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)

12) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{3}{\sqrt{2}+\sqrt{3}}+\dfrac{-1}{\sqrt{2}-1}\)

Giúp em giải với ạ! Help me~!

\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\)

\(2\sqrt{5}-3\sqrt{45}+\sqrt{500}\)

\(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}\)

\(\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}\)

\(\sqrt{3}-\sqrt{4+2\sqrt{3}}\)

\(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}\)

\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}\)

Thực hiện phép tính :

1 ) \(\frac{1}{5+2\sqrt{6}}\)\(+\)\(\frac{1}{5-2\sqrt{6}}\)

2 ) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}\)\(+\)\(\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

3 ) \(\frac{1}{2\sqrt{5}+3\sqrt{2}}\)\(-\)\(\frac{1}{2\sqrt{5}-3\sqrt{2}}\)

4 ) \(\frac{1}{3\sqrt{2}-4}\)\(-\)\(\frac{1}{3\sqrt{2}+4}\)

5 ) \(\frac{1}{\sqrt{7-\sqrt{24}}+1}\)\(-\)\(\frac{1}{\sqrt{7-\sqrt{24}}-1}\)

6 ) \(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\)\(-\)\(\frac{1}{\sqrt{\sqrt{3}+1}+1}\)

7 ) \(\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)\(+\)\(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}\)

8 ) \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)\(+\)\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)

+Tuấn 10B_2 (T ko biết đánh word nên dùng tạm .V)

GPT: \(\(\sqrt{x+3}+\sqrt[3]{x}=3\)\) (Bài này cách lp 9 dễ t ko giải nữa)

Vì \(\(f\left(x\right)=\sqrt{x+3}+\sqrt[3]{x}=3\)\) là hàm tăng trên tập [-3;\(\(+\infty\)\))

Ta có: Nếu \(\(x>1\Leftrightarrow f\left(x\right)>f\left(1\right)=3\)\)nên pt vô nghiệm

Nếu \(\(-3\le x< 1\Leftrightarrow f\left(x\right)< f\left(1\right)=3\)\)nên pt vô nghuêmj

Vậy x = 1

B2, GHPT: \(\(\hept{\begin{cases}2x^2+3=\left(4x^2-2yx^2\right)\sqrt{3-2y}+\frac{4x^2+1}{x}\\\sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\end{cases}}\)\)

ĐK \(\(\hept{\begin{cases}-\frac{1}{2}\le y\le\frac{3}{2}\\x\ne0\\x\ne-\frac{1}{2}\end{cases}}\)\)

Xét pt (1) \(\(\Leftrightarrow2x^2+3-4x-\frac{1}{x}=x^2\left(4-2y\right)\sqrt{3-2y}\)\)

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

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

Xét hàm số \(\(f\left(t\right)=t^3+t\)\)trên R có \(\(f'\left(t\right)=3t^2+1>0\forall t\in R\)\)

Suy ra f(t) đồng biến trên R . Nên \(\(f\left(-\frac{1}{x}+1\right)=f\left(\sqrt{3-2y}\right)\Leftrightarrow-\frac{1}{x}+1=\sqrt{3-2y}\)\)

Thay vào (2) \(\(\sqrt{2-\left(1-\frac{1}{x}\right)}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\sqrt{\frac{1}{x}+1}=\frac{\sqrt[3]{x^2\left(x+2\right)}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\left(2x+1\right)\sqrt{\frac{1}{x}+1}=x+2+\sqrt[3]{x^2\left(x+2\right)}\)\)

\(\(\Leftrightarrow\left(2+\frac{1}{x}\right)\sqrt{1+\frac{1}{x}}=1+\frac{2}{x}+\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow f\left(\sqrt{1+\frac{1}{x}}\right)=f\left(\sqrt[3]{1+\frac{2}{x}}\right)\)\)

\(\(\Leftrightarrow\sqrt{1+\frac{1}{x}}=\sqrt[3]{1+\frac{2}{x}}\)\)

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

Đặt \(\(\frac{1}{x}=a\)\)

\(\(\Rightarrow Pt:\left(a+1\right)^3=\left(2a+1\right)^2\)\)

Tự làm nốt , mai ra lớp t giảng lại cho ...

Mọi người ơi giúp em bài này với

Ai đã từng học lớp 9 rồi thì giúp em pls

Tính:

1, \(\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}-1}\)

2, \(\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}\)

3, \(\frac{1}{3+\sqrt{5}}+\frac{1}{\sqrt{5}-3}\)

4, \(\frac{1}{\sqrt{2}-\sqrt{6}}-\frac{1}{\sqrt{6}+\sqrt{2}}\)

5, \(\frac{1}{3-2\sqrt{2}}+\frac{1}{2+\sqrt{5}}\)

6, \(\frac{1}{3-2\sqrt{2}}-\frac{1}{3+2\sqrt{2}}\)

7, \(\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{2}{1-\sqrt{7}}\)

8, \(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{3}{18+2\sqrt{3}}\)

9, \(\frac{1}{5+2\sqrt{6}}-\frac{1}{5-2\sqrt{6}}\)

10, \(\frac{4}{1-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\)