Tìm x :
a) 4x2 - 9 = 0
b) 2x2 + 0,36 = 1
c) \(\frac{5}{12}.\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
d) 3x2 + 7 = -4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LD
19 tháng 6 2019
a.
\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{9}-\sqrt{7}}{9-7}\)
\(=\frac{\sqrt{9}-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{1}}{2}\)
\(=\frac{3-1}{2}=1\)
b.
\(B=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
c.
\(C=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{6-1}+\frac{4\sqrt{6}+8}{6-4}-\frac{36+12\sqrt{6}}{9-6}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
20 tháng 8 2019
d)D=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)( \(x\ge2\))
=\(\sqrt{x+2\sqrt{2}.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}\)
=\(\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}\)
=\(\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
=\(\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)(1)
TH1: \(2\le x\le4\)
Từ (1)<=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}\)
=\(2\sqrt{2}\)
TH2. x\(>4\)
Từ (1) <=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{2}+\sqrt{x-2}\)=\(2\sqrt{x-2}\)
Vậy \(\left[{}\begin{matrix}2\le x\le4\\x>4\end{matrix}\right.< =>\left[{}\begin{matrix}D=2\sqrt{2}\\D=2\sqrt{x-2}\end{matrix}\right.\)
CW
23 tháng 7 2016
a) \(\frac{6^7}{4^3\cdot9^2}=\frac{2^7\cdot3^7}{2^6\cdot3^4}=2\cdot3^3=2\cdot27=54\)
b) \(\frac{12^3\cdot15^3}{4^3\cdot25^2\cdot9^2}=\frac{2^6\cdot3^3\cdot3^3\cdot5^3}{2^6\cdot5^4\cdot3^4}=\frac{3^2}{5}=1,8\)
c) \(\frac{2^{11}+3\cdot2^{10}}{10\cdot4^5}=\frac{2^{10}\left(2+3\right)}{2\cdot5\cdot2^{10}}=\frac{1}{2}=0,5\)
d) \(\frac{3^8\cdot2-3^6}{2\cdot17\cdot3^7}=\frac{3^6\left(3^2\cdot2-1\right)}{2\cdot17\cdot3^7}=\frac{1}{2\cdot3}=\frac{1}{6}\)
9 tháng 3 2023
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
a, 4x2 - 9 = 0 => (2x)2 = 9 => 2x = 3 hoặc 2x = -3 => x = 3/2 hoặc x = -3/2
b, 2x2 + 0,36 = 1 => 2x2 = 0,64 => x2 = 0,32 = 8/25 => \(\orbr{\begin{cases}x=\sqrt{\frac{8}{25}}\\x=-\sqrt{\frac{8}{25}}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2\sqrt{2}}{5}\\x=\frac{-2\sqrt{2}}{5}\end{cases}}\)
c, \(\frac{5}{12}.\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
\(\Rightarrow\frac{5}{12}.\sqrt{x}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\div\frac{5}{12}\)
\(\Rightarrow\sqrt{x}=\frac{6}{5}\)
\(\Rightarrow x=\left(\frac{6}{5}\right)^2=\frac{36}{25}\)
d, 3x2 + 7 = -4 => 3x2 = -4 - 7 => 3x2 = -11 => x2 = -11/3 (vô lý) => x ∈ Ø