ĐKXĐ: x 0
√x - √(4x) + √(9x) = 6
√x - 2√x + 3√x = 6
2√x = 6
√x = 6 : 2
√x = 3
x = 9 (nhận)
Vậy x = 9
ĐKXĐ: x 0
√x - √(4x) + √(9x) = 6
√x - 2√x + 3√x = 6
2√x = 6
√x = 6 : 2
√x = 3
x = 9 (nhận)
Vậy x = 9
Câu 2: Tìm x biết:
a. \(\sqrt{x-1}=2\)
b. \(\sqrt{3x+1}=\sqrt{4x-3}\)
c. \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
d. \(\sqrt{x^2-4x+4}=\sqrt{6+2\sqrt{5}}\)
tìm x biết a,\(\sqrt{x^2-4x+4}=7\) b,\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\sqrt{9x+27}=6\)
Tìm x biết:
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+15}=6\)
ĐKXĐ:\(x\ge-5\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\) tìm x
GIẢI PHƯƠNG TRÌNH
a) \(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\)
b) \(\sqrt{9x^2+12x+4}=4x\)
c) \(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
d) \(\sqrt{5x-6}-3=0\)
Tìm x
a. \(\sqrt{4x}< =10\) b. \(\sqrt{9x}>=3\) c. \(\sqrt{4x^2+4x+1}=6\) d. \(\sqrt{9x-9}-2\sqrt{x-1}=6\)
e. \(\sqrt{4x^2-4x+1}=x-1\) f. \(\sqrt{2x+1}=\sqrt{x-1}\) g. \(\sqrt{x^2-x-1}=\sqrt{x-1}\)
h. \(\sqrt{9x^2+6x+1}=\sqrt{12-6\sqrt{3}}\)
a)\(\sqrt{4-5x}=12\) tìm x
b)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
c)\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
2) giải pt
3) \(\sqrt{4x+1}=x+1\)
4) \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
5) \(\sqrt{4x^2-12x+9}=7\)
6) \(5\sqrt{9x-9}-\sqrt{4x-4}-\sqrt{x-1}=36\)
giúp mk vs ah