Tim:m
m+m X\(\frac{1}{3}\):\(\frac{4}{48}\)+m:\(\frac{4}{14}\)=330
Tìm x,y,z khi:
1,\(\frac{x}{7}=\frac{y}{3}vàx-24=y\)
2,\(\frac{x}{5}=\frac{y}{7}=\frac{z}{2}và,y-x=48\)
3,\(\frac{x-1}{2005}=\frac{3-y}{2006}và,x-y=4009\)
4,\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}vã-y-z=28\)
5,\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}và2x+3y-z=-14\)
6,\(3x=y;5y=4zvà6x+7y+8z\)
Tìm tập hợp các số nguyên x thỏa mãn :
a, \(3\frac{1}{3}:2\frac{1}{2}-1< x< 7\frac{2}{3}.\frac{3}{7}+\frac{5}{2}\)
b,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
Bài 14: Tính x biết :
m) \(\frac{3}{4}-x=\frac{1}{5}\) o) \(\frac{3}{5}x+\frac{1}{4}=\frac{1}{10}\) p) \(-\frac{4}{3}x+\frac{3}{2}=\frac{5}{6}\)
m) \(\frac{3}{4}-x=\frac{1}{5}\)
\(x=\frac{3}{4}-\frac{1}{5}\)
\(x=\frac{11}{20}\)
Vậy \(x=\frac{11}{20}\)
o) \(\frac{3}{5}x+\frac{1}{4}=\frac{1}{10}\)
\(\frac{3}{5}x=\frac{1}{10}-\frac{1}{4}\)
\(\frac{3}{5}x=-\frac{3}{20}\)
\(x=\left(-\frac{3}{20}\right):\frac{3}{5}\)
\(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}\)
p) \(-\frac{4}{3}x+\frac{3}{2}=\frac{5}{6}\)
\(-\frac{4}{3}x=\frac{5}{6}-\frac{3}{2}\)
\(-\frac{4}{3}x=-\frac{2}{3}\)
\(x=\left(-\frac{2}{3}\right):\left(-\frac{4}{3}\right)\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
m) \(\frac{3}{4}\) - x = \(\frac{1}{5}\)
x= \(\frac{3}{4}\) - \(\frac{1}{5}\)
x= \(\frac{11}{20}\)
o) \(\frac{3}{5}\)x + \(\frac{1}{4}\) = \(\frac{1}{10}\)
\(\frac{3}{5}\)x = \(\frac{1}{10}\) - \(\frac{1}{4}\)
\(\frac{3}{5}\)x = -\(\frac{3}{20}\)
x = \(\frac{-3}{20}\):\(\frac{3}{5}\)
x = \(\frac{-1}{4}\)
p) \(\frac{-4}{3}\)x + \(\frac{3}{2}\) = \(\frac{5}{6}\)
\(\frac{-4}{3}\)x = \(\frac{5}{6}\) - \(\frac{3}{2}\)
\(\frac{-4}{3}\)x = \(\frac{-2}{3}\)
x = \(\frac{-2}{3}\) : \(\frac{-4}{3}\)
x = \(\frac{1}{2}\)
CHÚC BẠN HỌC TỐT
Bài 14: Tính x biết :
k) \(\frac{3}{4}-x=\frac{1}{5}\) l) \(\frac{7}{2}+2.x=\frac{-3}{4}\) m) \(\frac{3}{4}-x=\frac{1}{5}\) n) \(\frac{1}{3}+\frac{2}{3}.x=\frac{4}{3}\)
1) \(\frac{3}{4}-x=\frac{1}{5}\)<=> x=\(\frac{3}{4}-\frac{1}{5}=\frac{11}{20}\)
2) \(\frac{7}{2}+2x=-\frac{3}{4}\)<=> 2x=\(-\frac{3}{4}-\frac{7}{2}=-\frac{17}{4}\)<=> \(x=-\frac{17}{8}\)
3) \(\frac{1}{3}+\frac{2x}{3}=\frac{4}{3}\)<=> 1+2x=4<=> 2x=3<=> x=3/2
1) \(\frac{3}{4}-x=\frac{1}{5}\)
\(x=\frac{3}{4}-\frac{1}{5}\)
\(x=\frac{11}{20}\)
2)\(\frac{7}{2}+2x=\frac{-3}{4}\)
\(2x=\frac{-3}{4}-\frac{7}{2}\)
\(2x=-\frac{17}{4}\)
\(x=-\frac{17}{4}:2\)
\(x=-\frac{17}{8}\)
3)\(\frac{1}{3}+\frac{2}{3}x=\frac{4}{3}\)
\(\frac{2}{3}x=\frac{4}{3}-\frac{1}{3}\)
\(\frac{2}{3}x=1\)
\(x=1:\frac{2}{3}\)
\(x=\frac{3}{2}\)
^...^
^_^
k) \(\frac{3}{4}\) - x = \(\frac{1}{5}\)
x = \(\frac{3}{4}\) - \(\frac{1}{5}\)
x = \(\frac{11}{20}\)
l) \(\frac{7}{2}\) + 2.x = \(\frac{-3}{4}\)
\(\frac{7}{2}\) + 2x = \(\frac{-3}{4}\)
2x = \(\frac{-3}{4}\) -\(\frac{7}{2}\)
2x = \(\frac{-17}{4}\)
x = \(\frac{-17}{4}\) : 2
x = \(\frac{-17}{8}\)
m) Lặp lại câu k
n) \(\frac{1}{3}\) + \(\frac{2}{3}\).x = \(\frac{4}{3}\)
\(\frac{1}{3}\) + \(\frac{2}{3}\)x = \(\frac{4}{3}\)
\(\frac{2}{3}\)x = \(\frac{4}{3}\) - \(\frac{1}{3}\)
\(\frac{2}{3}\)x = 1
x = 1 : \(\frac{2}{3}\)
x = \(\frac{3}{2}\)
CHÚC BẠN HỌC TỐT
g*) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và x - 2y + 3z = 14
h*) \(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và x +y + z = 48
g*) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và x - 2y + 3z = 14
h*) \(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và x +y + z = 48
g) Đặt k = \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\)
=> \(\begin{cases}x-1=2k\\y-2=3k\\z-3=4k\end{cases}\)
=> \(\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}\)
=> x - 2y + 3z = 2k+1 - 6k - 4 + 12k + 9 = 8k + 6
=> 8k + 6 = 14
=> k = 1
=> \(\begin{cases}x=2\\y=5\\z=7\end{cases}\)
Nguyễn Huy Thắng Hoàng Lê Bảo Ngọc giúp câu h với
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)
\(\left[\begin{array}{nghiempt}12x-15y=0\\20z-12x=0\\15y-20z=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}12x=15y\\20z=12x\\15y=20z\end{array}\right.\)
\(\left[\begin{array}{nghiempt}\frac{x}{15}=\frac{y}{12}\\\frac{z}{12}=\frac{x}{20}\\\frac{y}{20}=\frac{z}{15}\end{array}\right.\)
TÌM x:
a,(5-x)+12=-25
b,12-4.(x-2)=-4
c,-15-/3-x/=-19
d,\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2.x\right)=-4\)
e,\(\left(x+\frac{1}{5}\right)m\text{ũ}2+\frac{17}{25}=\frac{26}{25}\)\
f,\(\frac{x}{3}+\frac{x}{7}=\frac{1}{7}+\frac{3}{14}\)
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
a) (5 - x) +12 = -25
<-> 5 - x + 12 = -25
<-> 17 - x = - 25
<-> x = 42
b) 12 - 4(x - 2) = -4
<-> 12 - 4x + 8 = -4
<-> 20 - 4x = -4
<-> 4x = 24
<-> x = 6
a) (5 - x) + 12 = -25
<=> -x = -25 - 12 - 5
<=> -x = -42
<=> x = 42
b) 12 - 4(x - 2) = -4
<=> 12 - 4x + 8 = -4
<=> -4x = -4 - 8 - 12
<=> -4x = -24
<=> x = 6
c) -15 - |3 - x| = -19
<=> -|3 - x| = -4
<=> 3 - x = 4 hoặc 3 - x = -4
<=> x = -1 hoặc x = 7
1.tìm x
a.\(\frac{-5}{8}+x=\frac{4}{9}\)
b.\(1^3_4.x+1^1_2=-\frac{4}{5}\)
c.\(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)
d.\(x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
e.\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
f.\(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
g.\(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
Bài 1:
a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)
hay \(x=\frac{77}{72}\)
Vậy: \(x=\frac{77}{72}\)
b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)
\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)
hay \(x=-\frac{46}{35}\)
Vậy: \(x=-\frac{46}{35}\)
c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)
\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)
hay \(x=\frac{2}{3}\)
Vậy: \(x=\frac{2}{3}\)
d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)
\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)
hay \(x=\frac{25}{42}\)
Vậy: \(x=\frac{25}{42}\)
e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)
\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)
\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)
hay \(x=-\frac{4}{5}\)
Vậy: \(x=-\frac{4}{5}\)
f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)
\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)
\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)
Vậy: \(x=-\frac{2}{3}\)
g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)
bài 2 tìm x thuộc Z
a) \(\frac{-1}{5}< \frac{1}{x}< \frac{1}{7}\) b) \(\frac{14}{5}< \frac{x}{5}< 4\)
c)\(\frac{1}{3}< \frac{9}{x}< \frac{1}{2}\)
a) \(\dfrac{-1}{5}< \dfrac{1}{x}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{1}{-5}< \dfrac{1}{x}< \dfrac{1}{7}\)
\(\Rightarrow\) \(x< -5\) và \(x< 7\left(x\in Z\right)\)
b) \(\dfrac{14}{5}< \dfrac{x}{5}< 4\)
\(\Leftrightarrow\dfrac{14}{5}< \dfrac{x}{5}< \dfrac{20}{5}\)
\(\Rightarrow14< x< 20\) mà \(x\in Z\)
\(\Rightarrow x\in\left\{15;16;17;18;19\right\}\)
c) \(\dfrac{1}{3}< \dfrac{9}{x}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{9}{27}< \dfrac{9}{x}< \dfrac{9}{18}\)
\(\Rightarrow18< x< 27\) mà \(x\in Z\)
\(\Rightarrow x\in\left\{19;20;...;25;26\right\}\)