Lời giải:
ĐKXĐ: $x\geq 1$
Áp dụng BĐT AM-GM:
$x^2+4=3x+2\sqrt{x-1}\leq 3x+(x-1)+1$
$\Leftrightarrow x^2+4\leq 4x$
$\Leftrightarrow (x-2)^2\leq 0$
Mà $(x-2)^2\geq 0$ với mọi $x\geq 1$
Do đó $(x-2)^2=0$
$\Rightarrow x=2$ (thỏa mãn)
giai pt \(x^2-3x=2\sqrt{x-1}-4\)
giai p.t : \(x\sqrt{x^2-x+1}+4\sqrt{3x+1}=x^2+x+3\)
giai pt sau
\(\sqrt{3x-1}-\sqrt{x+2}.\sqrt{3x^2+7x+2}+4=4x-2\)
\(x^2-5x+3.\sqrt{2x-1}=2.\sqrt{14-2x}+5\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
giai phuong trinh
a) \(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
b) x2 - 25 = y ( y + 6 ) (x; y nguyên)
giai
\(4\sqrt{x+3}+\sqrt{19-3x}=x^2+2x+9\)
\(\hept{\begin{cases}x^3+xy^2=y^6+y^4\\2\sqrt{y^4+1}+\frac{1}{x^2+1}=3-4x^3\end{cases}}\)
giai pt
a,(x-2)4+(x-2)(5x2-14x+13)+1=0
b,(x2-x)2-2x(3x-5)-3=0
c,x4+4x3+4x+1=0
d,x4+x3+x+1=0
giai pt
\(\frac{1}{\left(2x+1\right)^2}+\frac{1}{\left(3x+1\right)^2}=\frac{5}{4\left(x+2\right)^2}\)
Giai phương trình \(3\sqrt{x+4}+3\sqrt{1-x}+4\sqrt{3x+9}=x^2+7x+21\)
Giai phuong trinh \(x\sqrt{x^2-x+1}+2\sqrt{3x+1}=x^2+x+3\)
Giai pt sau:
1/x^2-3x+2 +1/x^2-5x+6 +1/x^2-7x+12 =2(Tất cả =2 nhé!)
