hộ e vs ak

Giải các pt vô tỉ sau ( bằng phương pháp đặt ẩn phụ đưa về phương trình tích )

a) \(\sqrt{x^3+x^2+3x+3}+\sqrt{2x}=\sqrt{x^2+3}+\sqrt{2x^2+2x}\)

b) \(\sqrt{x^2-3x}+2\sqrt{x}-4\sqrt{x-3}-x+8=0\)

c) \(\left(5x^2+4x+3\right)\sqrt{x}=\left(x+3\right)\sqrt{5x^2+4x}\)

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

e)\(\left(x^2+2x+1\right)3\sqrt{x^2+\frac{3}{x}}=x^3+2x^2+5\)

giải phương trình :

1,        \(\sqrt{4-x^2}+2\sqrt[3]{x^4-4x^3+4x^2}=\left(x-1\right)^2+1-\left|x\right|\)

2,        \(2x^3+9x^2-6x\left(1+2\sqrt{6x-1}\right)+2\sqrt{6x-1}+8=0\)

3,         \(x^3-3x+1=\sqrt{8-3x^2}\)

4,         \(\left(4x^2+x-1\right)\sqrt{x^2+x+2}=\left(4x^2+3x+5\right)\sqrt{x^2-1}\)

5,          \(\sqrt[3]{3-x^3}=2x^3+x-3\)

6,          \(\sqrt[3]{x^2+3x+3}+\sqrt[3]{2x^2+3x+2}=6x^2+12x+8\)

7,          \(\frac{x^2+2x-8}{x^2-2x+3}=\left(x+1\right)\left(\sqrt{x+2}-2\right)\)

8,          \(\frac{4x-1}{\sqrt{4x-3}}+\frac{11-2x}{\sqrt{5-x}}=\frac{15}{2}\)

9,          \(x^2-4x+14+\sqrt{x+4}=2\sqrt{1+12x}+\sqrt{1+\sqrt{1+12x}}\)

a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\)      ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ;  2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)

Giải phương trình:

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

2, \(x^3-3x^2+2\sqrt{\left(x+2\right)^3}-6x=0\)

3, \(2x^3-x^2-3x+1=\sqrt{x^5+x^4+1}\)

4, \(5\sqrt{x^4+8x}=4x^2+8\)

5, \(\left(x^2+4\right)\sqrt{2x+4}=3x^2+6x-4\)

6, \(\left(x^2-6x+11\right)\sqrt{x^2-x+1}=2\left(x^2-4x+7\right)\sqrt{x-2}\)

Giải phương trình:

a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)

b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)

c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)

d) \(\sqrt{x^2-6x+9}+x=11\)

e) \(\sqrt{3x^2-4x+3}=1-2x\)

f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

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

b)\(x^2+x+12\sqrt{x+1}=36\)

c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)

e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)

f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)

g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)

h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)

i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)

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

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

Giải các pt sau:

a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)

b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)

c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)

d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)

f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)

g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)

Giải phương trình vô tỉ:

a)\(\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}\)

b)\(2\sqrt{x+3}=9x^2-x-4\)

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

d)\(\sqrt{x-2+\sqrt{2x+5}}+\sqrt{x+2+3\sqrt{2x+5}}=7\sqrt{2}\)

e)\(\sqrt{4x+5}=2x^2-6x-1\)

f)\(\sqrt{5+\sqrt{x-1}}=6-x\)

g)\(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\)

Tìm ĐKXĐ giúp mik luôn nha!