căn bậc 3 (7x-6)=2
lập 2 vế có
7x-6=8
=>7x=14
=>x=2
căn bậc 3 (7x-6)=2
lập 2 vế có
7x-6=8
=>7x=14
=>x=2
1)\(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
2)\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)
3)\(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
4)\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
5)\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)
6)\(2\left(5x-3\right)\sqrt{x+1}+\left(x+1\right)\sqrt{3-x}=3\left(5x+1\right)\)
7)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
giải pt sau:
\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{7x+7}\sqrt{7x-6}=181-14x\)
Tìm `ĐKXĐ`:
\(\sqrt{\dfrac{-5}{6+x}}\)
\(\sqrt{\dfrac{-2}{6-x}}\)
\(\sqrt{\dfrac{-x+3}{-6}}\)
\(\sqrt{\dfrac{7x-1}{-9}}\)
\(\sqrt{\dfrac{x+2}{x^2+2x+1}}\)
\(\sqrt{\dfrac{x-2}{x^2-2x+4}}\)
Giải phương trình
a) \(\sqrt{x-2}=\sqrt{x^2-4x+3}\)
b) \(2\left(\sqrt{\dfrac{x-1}{4}}-3\right)=2\sqrt{\dfrac{4x-4}{9}}-\dfrac{1}{3}\)
c) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
d) \(4+\sqrt{2x+6-6\sqrt{2x-3}}=\sqrt{2x-2+2\sqrt{2x-3}}\)
GPT:
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}-3\sqrt{3}x=6\sqrt{3}\)
Gpt: \(5x^2+3x+6=\left(7x+1\right)\sqrt{x^2+3}\)
Giải pt sau đây
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}-3\sqrt{3}x=6\sqrt{3}\)
Giải phương trình
\(\sqrt{x^2+8}-7x=\sqrt{x^2+3}-6\)
giải pt \(\sqrt{3x^2-6x-6}=3\sqrt{\left(2-x\right)^5}+\left(7x-19\right)\sqrt{2-x}\)