a, 2x+ 1/4 = 3/2
b, (x-5) - 1/3 = 2/5
c, (4,5-2x):3/4=4/3
Tìm x:
a) 2/5 = 3/4 : x = -1/2
b) 5/7 - 2/3 . x = 4/5
c) 1/2x + 3/5x = -2/3
d) 4/7x - x = -9/14
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
Tìm x
a) -3 1/2 : (4/5-1/2x) = 2^2
b) 2x + 3x = 5
c) -2/3x - 1/3x = -2
d) -2/3 (x+1) - 1/2 = -1/3
a: =>-7/2:(4/5-1/2x)=4
=>4/5-1/2x=-7/2:4=-7/8
=>1/2x=4/5+7/8=67/40
=>x=67/20
b: =>5x=5
=>x=1
c: =>x(-2/3-1/3)=-2
=>-x=-2
=>x=2
d: =>-2/3(x+1)=-1/3+1/2=1/6
=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4
=>x=-1/4-1=-5/4
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a)|-x+2/5|+1/2=3,5 b)21/5+3:|x/4-2/3|=6
c)7,5-3|5-2x|=-4,5 d)1/3-|5/4-2x|=1/4
e)21/5+3:|x/4-2/3|=6
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
3x^4 + 3x^2y^2 + 6x^3y - 27x^2
x^4 + x^3 - x^2 + x
2x^5 - 6x^4 - 2a^2x^3 - 6ax^3
x^5 + x^4 + x^3 + x^2 + x + 1
x^3 - 1 + 5x^2 - 5 + 3x - 3
1/4.(a + 1)^2 - 4/9.(a - 2)^2
12a^2b^2 - 3.(a^2b^2)^2
4x^2y^2 - (x^2 + y^2 - a^2)^2
(a + b + c)^2 + (a + b - c)^2 - 4c^2
x^3 - 1 + 5x^2 - 5 + 3x - 3
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
Tìm x biết
a) 4/5 + x = 2/3
b)3/4 - x = 1/3
c) x - 5/9 = -2/3
d)(4,5 - 2x) : 3/4= 1 1/3
a) Ta có: \(\frac{4}{5}+x=\frac{2}{3}\)
hay \(x=\frac{2}{3}-\frac{4}{5}=\frac{10}{15}-\frac{12}{15}=\frac{-2}{15}\)
Vậy: \(x=\frac{-2}{15}\)
b) Ta có: \(\frac{3}{4}-x=\frac{1}{3}\)
hay \(x=\frac{3}{4}-\frac{1}{3}=\frac{9}{12}-\frac{4}{12}=\frac{5}{12}\)
Vậy: \(x=\frac{5}{12}\)
c) Ta có: \(x-\frac{5}{9}=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+\frac{5}{9}=\frac{-6}{9}+\frac{5}{9}=-\frac{1}{9}\)
Vậy: \(x=\frac{-1}{9}\)
d) Ta có: \(\frac{\left(4,5-2x\right)}{\frac{3}{4}}=1\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{2}-2x=\frac{4}{3}\cdot\frac{3}{4}=1\)
\(\Leftrightarrow2x=\frac{9}{2}-1=\frac{7}{2}\)
hay \(x=\frac{7}{4}\)
Vậy: \(x=\frac{7}{4}\)
Thực hiện phép tính ( theo hằng đẳng thức đáng nhớ )
1. ( x - 2 ) mũ 2 - ( x + 3 ) mũ 2 + ( x+ 4 ) ( x-4 )
2. 2 ( 3x - 2 ) mũ 2 - 3 ( 2x + 5 ) mũ 2 - 6 ( x -1 ) ( x+ 1 )
3. ( 2x + 3 ) mũ 2 - 2 ( 2x + 3 ) ( 2x + 5 ) + ( 2x + 5 ) mũ 2
4. ( a + 1 ) ( a + 2 ) ( a mũ 2 + 4 ) ( a - 1 ) ( a mũ 2 + 1 ) ( a - 2 )
5. ( a + 2b - 3c - d ) ( a + 2b + 3c + d ) - ( a mũ 2 + 4ab + 4b mũ 2 )
Các bạn có thể giúp mik đc ko, mik cảm ơn các bạn nhiều !!!!
a/Thực hiện phép tính: (1−1/2+1/3).(4/5−3/4)2
b/Tìm x biết: (2x-1)4=81
b: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)