\(\sqrt{3x-10}=\sqrt{x-4}\left(đk:x\ge4\right)\)
\(\Leftrightarrow3x-10=x-4\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(ktm\right)\)
Vậy \(S=\varnothing\)
bài 1
a) \(\sqrt{2X+1}\)
b)\(\sqrt{x^2-4}\)
c) \(\dfrac{3}{\sqrt{3X+5}}\)
d) \(\sqrt{X-3}-\sqrt{10-x}\)
e) \(\sqrt{x+4}+\dfrac{2-X}{x^2-16}\)
Tìm ĐK để căn thức sau xác định:
a) \(\sqrt{x^2+3x-10}\)
b) \(\sqrt{\dfrac{4x-4-x^2}{5}}\)
c) \(\sqrt{x-4\sqrt{x-4}}\)
Tìm x biết:
a.\(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)
b.\(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)
c.\(\sqrt{\left(x-2\right)^2}=10\)
d.\(\sqrt{9x^2-6x+1}=15\)
e.\(\sqrt{3x+4}=3x-8\)
a) \(\sqrt{x-3}-\sqrt{10-x}\)
b) \(\sqrt{x+4}+\dfrac{2-X}{x^2-16}\)
c) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-4}}\)
d) \(\dfrac{\sqrt{2x-1}}{3x+2}\)
e) \(\dfrac{-2}{\sqrt{x^2+2x+2}}\)
1. \(x^4-x^2+3x+5=2\sqrt{x+2}\)
2. \(\sqrt{x^2+x}+\sqrt{x-x^2}=2x+2\)
3. \(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
4. \(\sqrt{2x^2-1}+\sqrt{x^2-3x+2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)
Tìm x:
1, \(\sqrt{1-2x}+\sqrt{1+2x}=\sqrt{4+x}\)
2,\(\sqrt{3x+1}-\sqrt{2-x}=\sqrt{3x-2}\)
3, \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=4\)
4, \(3x^2-6x-4=4\left(x-1\right)\sqrt{3x+1}\)
1/
A = \(\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\) là một số nguyên
2/
a) Cho x = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\). Tính giá trị biểu thức:
P = \(\dfrac{x^4-4x^3+x^2+6x+12}{x^2-2x+12}\)
b) Cho x = \(1+\sqrt[3]{2}\) . Tính giá trị của biểu thức B = \(x^4-2x^4+x^3-3x^2+1942\)
3/
Rút gọn:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
B = \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
Làm ơn, giúp mik với. Mik đang cần gấp!
1)
2)
3)
4)
5)
a) cho x=\(1+\sqrt[3]{2}\) tính B = \(x^4-2x^5+x^3-3x^2+1942\)
b) cho x = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\) tính P =\(\dfrac{x^4-4x^3+x^2+6x+12}{x^2-2x+12}\)
c) cho x = \(1+\sqrt[3]{2}\)\(+\sqrt[3]{4}\) tính C = \(x^5-4x^4+x^3-x^2-2x+2015\)
