giúp mk vs : gpt :
A= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
GPT : a, \(\sqrt{x^2-2x+5}\) + \(2\sqrt{4x+5}\)= \(x^3\)-\(2x^2\)+ \(5x+4\)
b,\(2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\)
gpt \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
GPT: \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
GPT
A, X(X+5)=2\(\sqrt[3]{X^2+5X-2}\) -2
B,\(\sqrt[3]{1-X}\)+\(\sqrt{X+2}\)=1
C,13[(X^2-3X+6)^2+(X^2-2X+7)^2]=(5X^2-12X+33)^2
D,\(\sqrt{X+1}\)+\(\sqrt{X+8}\)=3X+\(\sqrt{2X}\)
E,\(\sqrt{X+2}\)+\(\sqrt{X^2+X+2}\)=2X+\(\sqrt{2X+1}\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
Gpt: \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
gpt:
\(x^4-2x^3+x=\sqrt{(x^2-x).2}\)
\(x(5x^3+2)-2(\sqrt{2x+1}-1)=0\)
\(\sqrt{x+\frac{3}{x}}=\frac{x^2-7}{2x+2}\)