Đặt \(\sqrt{5x^2+6x+5}=a,4x=b\left(a\ge0\right)\)
Khi đó Pt
<=> \(a\left(a^2+1\right)=b\left(b^2+1\right)\)
<=>\(\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
MÀ \(a^2+ab+b^2+1>0\)
=> \(a=b\)
=> \(\sqrt{5x^2+6x+5}=4x\)
=> \(\hept{\begin{cases}x\ge0\\11x^2-6x-5=0\end{cases}}\)
=>\(x=1\)
Vậy x=1
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}\)
b) \(\sqrt{\sqrt{6x}-4x}\)
c) \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}\)
d) \(\sqrt{\left(x-6\right)^6}\)
e) \(\sqrt{-12x+5}\)
f) \(2-4\sqrt{5x+8}\)
g) \(\sqrt{x^2-9}\)
\(\left(6x^2-5x+1\right)\left(x^2-5x+6\right)=4x^2\)
help me now
\(\left(x-x^2\right)\left(\sqrt{x-2}+2\right)=2x^3-5x^2+5x-2\)
\(\sqrt{2x-3+\sqrt{4x-7}}+\sqrt{2x+9+5\sqrt{4x-7}}=4\sqrt{2}\)
\(\left(\sqrt{3x+1}-\sqrt{x+2}\right)\left(\sqrt{3x^2+7x+2}+9\right)=6x-3\)
a, giải phương trình sau: \(4x^3+4x^2-5x+9=4\sqrt[4]{16x+8}\)
b, chứng minh phương trình sau vô nghiệm trên tập hợp số thực:
\(9x^4+x\left(12x^2+6x-1\right)+\left(x+1\right)\left(9x^2+12x+5\right)+1=0\)
giải phương trình \(x^3-5x^2+4x-5=\left(1-2x\right)\sqrt[3]{6x^2-5x+7}\)
Mn giúp mình vs
1, \(x^3-6x^2+10x-4=0\)
2, \(x^3+2x^2+2\sqrt{2}x+2\sqrt{2}=0\)
3, \(x^4+x^2-\sqrt{2}x+2=0
\)
4, \(x^4+5x^3-12x^2+5x+1=0\)
5, \(\left(x+5\right)\left(2x+12\right)\left(2x+20\right)\left(x+12\right)=3x^2\)
6, \(\left(x^2-5x+1\right)\left(x^2-4\right)=6\left(x-1\right)^2\)
7, \(x^4-9x^3+16x^2+18x+4=0\)
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)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
