\(\frac{2xX}{3}\)- \(\frac{1}{4}\): ( \(\frac{3}{2}\)-\(\frac{3}{4}\)) = \(\frac{1}{4}\)X\(\frac{2}{3}\)
Tìm x
a \(\frac{3}{4}+\frac{1}{4}xX=2\)
b X - \(\frac{2}{3}x\frac{9}{4}=2,5-\frac{1}{2}\)
c \(2xX+\frac{1}{3}=\frac{4}{3}\)
d \(X:\frac{2}{3}+0,75=\frac{9}{4}+3\)
e \(\frac{6}{4}:X+\frac{1}{2}=\frac{9}{4}x\frac{2}{3}\)
f \(Xx\frac{3}{5}-\frac{2}{5}=3\frac{3}{4}-1\)
a, 3/4 + 1/4.x=2
1/4.x = 2-3/4
1/4.x =5/4
x = 5/4:1/4
x = 5
b, x-2/3.9/4=2,5-1/2
x-2/3.9/4=2
x-2/3 =2:9/4
x-2/3 =8/9
x = 8/9+2/3
x = 14/9
c, 2.x+1/3=4/3
2.x =4/3-1/3
2.x =1
x =11:2
x = 1/2
Tìm x:
\(\frac{x}{5}=\frac{1}{2}-\frac{1}{5}\)
\(\frac{1}{3}+\frac{2}{3}xX=\frac{18}{21}\)
\(\frac{3}{4}xX+\frac{x}{6}=\frac{1}{6}\)
\(2xX+3\frac{1}{2}+x=24\frac{1}{4}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tìm X
\(\frac{1}{2+4}\)+\(\frac{1}{2+4+6}\)+\(\frac{1}{2+4+6+8}\)+........+\(\frac{1}{2+4+6+8+.....+2xX}\)=\(\frac{503}{1007}\)
Ai trong vòng tối nay giải được trước 9 giờ 3 kick
đặt 1/2 ra
phần mẫu áp dụng công thức tính tổng còn lại tự làm nhoa bé
\(4^x+\frac{1}{3}=\frac{3}{2}\)
\(\frac{1}{3}-\frac{2}{5}+\frac{3}{x}=\frac{3}{4}\)
\(2.\left(\frac{3}{4}-\frac{5}{x}\right)=\frac{4}{5}-3.x\)
\(\frac{3}{2}-4.\left(\frac{1}{4}-x\right)=\frac{2}{3}-\frac{7}{x}\)
\(3.2^x=4^5.16^2\)
Tìm x: \(\frac{3\frac{1}{2}x-4}{6}+\frac{2+\frac{1}{4}x}{3}=1\frac{1}{4}x\left(x-1\right)-\frac{7-\frac{3}{4}x}{3}\)