đk: x≥1
\(pt\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}+3\sqrt{x-1}=18\)
\(\Leftrightarrow6\sqrt{x-1}=18\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x-1=9\Leftrightarrow x=10\left(tm\right)\)
Vậy pt có nghiệm x = 10
1) Giải phương trình: a) \(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{91}}=0\) b) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\)
giải pt:
\(\dfrac{1}{2}\sqrt{x-2}-4\sqrt{\dfrac{4x-8}{9}}+\sqrt{9x+18}-5=0\)
Tìm x thỏa mãn:
\(\left(\sqrt{x}+1\right)\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
Cho bieu thuc \(\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)voi x\(\ge\)-1.Tim x sao cho bieu thuc co gia tri bang 18
Tìm x :
a, \(\sqrt{x^2-2x}=\sqrt{2-3x}\)
b, \(\sqrt{x-3}-2\sqrt{x^2-9}=0\)
c, \(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}\sqrt{9x-45}=4\)
d, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
e, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
f, \(\sqrt{x^2-4}-x+2=0\)
TÍNH: \(15\sqrt{x+1}-\sqrt{9x+9}=3\sqrt{4x+4}+12\)
a,\(\sqrt{x^2-6x+9}=3\) b,\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
giải phương trình
a, \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
b, \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
c, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
Giải phương trình
\(a.\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
\(b.\sqrt{3-x}-\sqrt{27-9x}+1,25.\sqrt{48-16x}=6\)
\(c.\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\dfrac{2}{7}\)
\(d.\sqrt{9x^2+12x+4}=4\)
\(\sqrt{9x-18}+5\sqrt{4x-8}=\sqrt{x-2}+3\)
