Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải phương trình
<=> |2x - 1| - 2 = x <=> |2x - 1| = x + 2
TH1: 2x - 1 = x + 2
Tự giải: x = 3
TH2: 1 - 2x = x + 2
Tự giải: x = -1/3
(Nhớ thêm điều kiện nhá)
......................?
mik ko biết
mong bn thông cảm
nha ................
1: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{\dfrac{10}{3}}\right)\cdot\left(\sqrt{\dfrac{6}{5}}+\sqrt{2}-\dfrac{4}{\sqrt{5}}\right)\)
\(=\left(\dfrac{4\sqrt{3}}{3}+\dfrac{3\sqrt{2}}{3}+\dfrac{\sqrt{30}}{3}\right)\cdot\left(\dfrac{\sqrt{30}}{5}+\dfrac{5\sqrt{2}}{5}-\dfrac{4\sqrt{5}}{5}\right)\)
\(=\dfrac{\left(4\sqrt{3}+3\sqrt{2}+\sqrt{30}\right)\left(\sqrt{30}+5\sqrt{2}-4\sqrt{5}\right)}{15}\)
2: \(=\left(2\sqrt{3}+6\sqrt{3}\right)\cdot\dfrac{\sqrt{3}}{2}-5\sqrt{6}\)
\(=\dfrac{8\sqrt{9}}{2}-5\sqrt{6}=4\sqrt{9}-5\sqrt{6}=12-5\sqrt{6}\)
Bài 4:
a: ĐKXĐ: x>=0; x<>1
b: \(P=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{2a^2+4}{-\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{-\sqrt{a}+1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{-2a^2-4}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{2}{a-1}\)
\(=\dfrac{-2a^2-4+2a^2+2a+2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{2a+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
Bài 1:
a: \(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+12\sqrt{5}=\sqrt{5}\)
b: \(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{8+2\sqrt{7}}\)
\(=2\sqrt{7}-1+\sqrt{7}+1=3\sqrt{7}\)
c:\(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\)
\(=\dfrac{1+\sqrt{3}-1+\sqrt{3}}{-2}=-\dfrac{2\sqrt{3}}{2}=-\sqrt{3}\)
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
1: =>|2x-1|=5
=>2x-1=5 hoặc 2x-1=-5
=>2x=6 hoặc 2x=-4
=>x=3 hoặc x=-2
2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=2\)
=>x-3=4
hay x=7
5: \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=>x-2=0 hoặc x+2=1
=>x=2 hoặc x=-1
Bài 1:
a, \(\sqrt{2x+5}=\sqrt{1-x}\)
\(\Rightarrow2x+5=1-x\Rightarrow2x+x=1-5\)
\(\Rightarrow3x=-4\Rightarrow x=-\dfrac{4}{3}\)
b, \(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Rightarrow x^2-x=3-x\)
\(\Rightarrow x^2-x+x=3\Rightarrow x^2=3\)
\(\Rightarrow x=\pm\sqrt{3}\)
c, \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\Rightarrow2x^2-3=4x-3\)
\(\Rightarrow2x^2-4x=0\Rightarrow2x.\left(x-2\right)=0\)
\(\Rightarrow x.\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Chúc bạn học tốt!!!
Sửa đề:
a, \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}=\dfrac{3+\sqrt{2}+3-\sqrt{2}}{9-2}=\dfrac{6}{7}\)
b, \(\sqrt{3}\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)\)
\(=\sqrt{3}.4\sqrt{3}=12\)
Câu 2
\(\sqrt{x-1}+\sqrt{4x-1}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)
Vậy...
Câu 1:
a) \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}=\dfrac{3-\sqrt{2}}{9-4}+\dfrac{3+\sqrt{2}}{9-4}=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{5}=\dfrac{6}{5}\)