ĐKXĐ: x>=2
\(\sqrt{9x-18}+5\sqrt{4x-8}=\sqrt{x-2}+3\)
=>\(3\sqrt{x-2}+10\sqrt{x-2}-\sqrt{x-2}=3\)
=>\(12\sqrt{x-2}=3\)
=>\(\sqrt{x-2}=\dfrac{1}{4}\)
=>x-2=1/16
=>x=33/16
Giải phương trình
\(a.\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
\(b.\sqrt{3-x}-\sqrt{27-9x}+1,25.\sqrt{48-16x}=6\)
\(c.\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\dfrac{2}{7}\)
\(d.\sqrt{9x^2+12x+4}=4\)
Giải PT:
a) -5x+7\(\sqrt{x}\) +12=0
b) \(\dfrac{1}{3}\)\(\sqrt{4x^2-20}\) +2\(\sqrt{\dfrac{x^2-5}{9}}\) -3\(\sqrt{x^2-5}=0\)
c) \(\sqrt{9x+27}+5\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=5\)
d) \(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=3\sqrt{x-2}+8\)
giải pt:
\(\dfrac{1}{2}\sqrt{x-2}-4\sqrt{\dfrac{4x-8}{9}}+\sqrt{9x+18}-5=0\)
Tìm x thỏa mãn:
\(\left(\sqrt{x}+1\right)\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
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, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
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, \(\sqrt{x^2-4}-x+2=0\)
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) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\)
a,\(\sqrt{x^2-6x+9}=3\) b,\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
Gỉai phương trình
a) \(\frac{3}{4}\sqrt{5x}-\sqrt{4x}+5=\frac{1}{4}\sqrt{4x}\)
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
c) \(\frac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\frac{2}{7}\)
d) \(\sqrt{3x^2+12x+4}=4\)
tìm x
1. 4x-\(\sqrt{9x^2+6x+1}\)=0
2. \(\sqrt{4x}+20-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=0\)
3. \(\sqrt{x^2}+x+1=x+2\)
4. \(\sqrt{6x^2}+2-\sqrt{3x^2}=\sqrt{26}-\sqrt{13}\)
5. \(\dfrac{\sqrt{x}-2}{\sqrt{x}-4}=\dfrac{\sqrt{x}-6}{\sqrt{x}-7}\)
6. \(\sqrt{4x^2-4x+1}=3\)
giải từng bước ra cho mik nha,thank mn
