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22 tháng 9 2017
Bài 1: a) 4 . ( \(\dfrac{1}{3}\) - x ) + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x
=> \(\dfrac{4}{3}\) - 4.x + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x
=> \(\dfrac{11}{6}\) = \(\dfrac{29}{6}\) . x
=> x = \(\dfrac{11}{29}\) .
b) \(\dfrac{5}{2}\) - 3 . ( \(\dfrac{1}{3}\) - x ) = \(\dfrac{1}{4}\) - 7.x
=> \(\dfrac{1}{4}\) - \(\dfrac{3}{2}\) = 7.x + 3.x
=> \(\dfrac{-5}{4}\) = 10.x
=> x = \(\dfrac{-1}{8}\).
TT
23 tháng 9 2017
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
19 tháng 5 2022
a: \(\left(2x-\dfrac{1}{2}\right)^3\)
\(=\left(2x\right)^3-3\cdot4x^2\cdot\dfrac{1}{2}+3\cdot2x\cdot\dfrac{1}{4}-\dfrac{1}{8}\)
\(=8x^3-6x^2+\dfrac{3}{2}x-\dfrac{1}{8}\)
b: \(\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)=\dfrac{1}{4}x^2-y^2\)
c: \(\left(x+\dfrac{1}{3}\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}+3\cdot x\cdot\dfrac{1}{9}+\dfrac{1}{27}\)
\(=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
d: \(\left(x-2\right)\left(x^2+2x+4\right)=x^3-8\)
30 tháng 6 2017
bạn sử dụng 7 hằng đẳnng thức đó
a. A – B)3 = A3 – 3A2B + 3AB2 – B3
b. A2 – B2 = (A – B)(A + B)
c. (A + B)3 = A3 + 3A2B + 3AB2 + B3
d. A3 – B3 = (A – B)(A2 + AB + B2)
CM
7 tháng 12 2017
\(\frac{2\frac{1}{2}x-1}{\frac{2}{3}}=\frac{\frac{-2}{3}}{1-2\frac{1}{2}x}\) ĐKXĐ \(x\ne\frac{2}{5}\)
\(\Leftrightarrow\)\(\frac{\frac{5}{2}x-1}{\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{5}{2}x-1}\)\(\Leftrightarrow\)\(\left(\frac{5}{2}x-1\right)^2=\frac{4}{9}\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+1=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+\frac{5}{9}=0\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-\frac{25}{6}x-\frac{5}{6}x+\frac{5}{9}=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{4}x^2-\frac{25}{6}x\right)-\left(\frac{5}{6}x-\frac{5}{9}\right)=0\)\(\Leftrightarrow\)\(\frac{25}{2}x\left(\frac{1}{2}x-\frac{1}{3}\right)-\frac{5}{3}\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{2}x-\frac{5}{3}\right)\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{2}{15}\end{cases}}\)
1 tháng 6 2022
b: \(\dfrac{x-1}{5}=\dfrac{2x+1}{3}\)
=>10x+5=3x-3
=>7x=-8
hay x=-8/7
c: \(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
=>259-7x=3x+39
=>-10x=-220
hay x=22
d: \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-1=x^2-4\)(vô lý)
e: \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
=>x+4=10 hoặc x+4=-10
=>x=6 hoặc x=-14
2 tháng 10 2017
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
25 tháng 10 2017
a,|x2−13x2−13| = 3232
b, 32−1232−12 ( 2x-1)=3434
c, |x-1|+2x=2
25 tháng 10 2017
a)\(\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{3}{2}\)
TH1
\(\dfrac{x}{2}-\dfrac{1}{3}=\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=\dfrac{11}{6}\)
=>x=\(\dfrac{11.2}{6}\)
=>x=\(\dfrac{11}{3}\)
TH2
\(\dfrac{x}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=-\dfrac{3}{2}+\dfrac{1}{2}\)
=>\(\dfrac{x}{2}=-1\)
=>x=-2
15 tháng 10 2018
Bài 1
a, \(D=1-\left|2x-3\right|\)
Ta có : \(\left|2x-3\right|\ge0\)
\(\Rightarrow1-\left|2x-3\right|\le1\)
Dấu "=" xảy ra khi \(\left|2x-3\right|=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=3:2=\dfrac{3}{2}\)
15 tháng 10 2018
\(b,\) Ta có : \(\left|10-5x\right|\ge0\Rightarrow\left|10-5x\right|+14,2\ge14,3\Rightarrow-\left|10-5x\right|-14,2\le-14,2\)
Dấu "=" xảy ra khi \(-\left|10-5x\right|=0\)
\(\Leftrightarrow10-5x=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy \(Emax=-14,2\Leftrightarrow x=2\)
\(c,\) Ta có : \(\left|5x-2\right|\ge0\)
\(\left|3y-12\right|\ge0\)
⇒ \(\left|5x-2\right|+\left|3y+12\right|-4\ge-4\)
⇒ \(4-\left|5x-2\right|-\left|3y+12\right|\le4\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2\\3y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
\(d,\) \(A=5-3\left(2x-1\right)^2\)
Ta có : \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2-5\ge-5\)
\(\Rightarrow5-3\left(2x-1\right)^2\le5\)
Dấu "=" xảy ra khi \(\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)
Từ pt=>3x-6=4x+2
=>4x-3x=-6-2
=>x=-8
Vậy x=-8