Question

[Phạm Kiến Kim Thùy]

Tìm x :

h/ \(\sqrt{x+5}-10=-4\)

i/ \(\sqrt{x-5}+2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)

j/ \(3\sqrt{2x}+\frac{1}{7}\sqrt{98x}-\sqrt{72x}+4=0\)

k/ \(\sqrt{4x^2-20}-\frac{1}{3}\sqrt{x^2-5}+\sqrt{\frac{9x^2-45}{16}}-\frac{1}{2}\sqrt{\frac{25x^2-125}{36}}=4\)

l/ \(\sqrt{4x+4}+\sqrt{9x+9}-\sqrt{x+1}=4\)

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

Giúp mk với nhé mn

Answer 1

[Akai Haruma]

k) ĐK: $x^2\geq 5$

PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$

$\Leftrightarrow 2\sqrt{x^2-5}=4$

$\Leftrightarrow \sqrt{x^2-5}=2$

$\Rightarrow x^2-5=4$

$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)

l) ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$

$\Leftrightarrow 4\sqrt{x+1}=4$

$\Leftrightarrow \sqrt{x+1}=1$

$\Rightarrow x+1=1$

$\Rightarrow x=0$

m) 

ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$

$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}=16$

$\Leftrightarrow \sqrt{x+1}=4$

$\Rightarrow x=15$ (thỏa mãn)

Answer 2

[Akai Haruma]

h) 

ĐKXĐ: $x\geq -5$

PT $\Leftrightarrow \sqrt{x+5}=6$

$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)

i) ĐKXĐ: $x\geq 5$

PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)

\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)

j) 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$

$\Leftrightarrow -2\sqrt{2x}+4=0$

$\Leftrightarrow \sqrt{2x}=2$

$\Rightarrow x=2$ (thỏa mãn)