\(\dfrac{x+ 3}{20}=\dfrac{5}{4}\)
\(\dfrac{2}{5}\) x \(\dfrac{3}{4}-\dfrac{1}{8}\) \(= \)
\(\dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\)
\(\dfrac{9}{20}-\dfrac{3}{5}\) x \(\dfrac{1}{4}\)\(= \)
\(\dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}\)\(=\)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
Tìm x, biết:
a) \(\dfrac{3}{4}x\) - \(\dfrac{7}{12}\) = \(\dfrac{5}{6}\) - \(\dfrac{2}{3}\)
b) -\(\dfrac{5}{x}\) = \(\dfrac{20}{28}\)
c) \(2\dfrac{1}{3}\) : x = 7
d) \(\dfrac{-105}{12}\) < x < \(\dfrac{20}{7}\)
\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)
\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)
hay x=1
b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)
c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)
d: \(\Leftrightarrow-8< x< 3\)
hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)
\(a)\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \dfrac{3}{4}x=\dfrac{2}{3}\\ x=\dfrac{2}{3}:\dfrac{3}{4}\\ x=\dfrac{8}{9}\\ b)-\dfrac{5}{x}=\dfrac{20}{28}\\ -5\cdot28=x\cdot20=-140\\ x=-140:20\\ x=-7\\ c)\dfrac{7}{3}:x=7\\ x=\dfrac{7}{3}:7\\ x=\dfrac{1}{3}\)
tìm x :
a, x . \(\dfrac{-5}{8}\) = \(\dfrac{15}{32}\) b, \(\dfrac{3}{10}\) : x =\(\dfrac{-9}{20}\)
c, \(\dfrac{-1}{4}\) x + \(\dfrac{4}{5}\) =\(\dfrac{3}{4}\) d, \(\dfrac{-7}{8}\) + \(\dfrac{2}{3}\) :x = \(\dfrac{3}{5}\) . \(\dfrac{-5}{12}\)
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)
\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)
\(\Leftrightarrow x=-\dfrac{3}{4}\)
\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)
\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{16}{15}\)
a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=-\dfrac{2}{3}\)
c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
#YVA6
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=\dfrac{-2}{3}\)
\(c,\dfrac{-1}{4}.x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)
\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
\(#yH\)
\(#NBaoNgoc\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
\(\)a) \(\dfrac{3}{5}+\dfrac{11}{20}\) b) \(\dfrac{5}{8}-\dfrac{4}{9}\) c) \(\dfrac{9}{16}\) x \(\dfrac{4}{3}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}\) e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}\)
a) \(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b) \(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c) \(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
a)\(=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b)\(=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c)\(=\dfrac{9\times4}{16\times3}=\dfrac{3}{4}\)
d)\(=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e)\(=\dfrac{3}{5}+\dfrac{4}{2}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
a) x + \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)
b) x - \(\dfrac{2}{5}\) = \(\dfrac{2}{7}\)
c) \(\dfrac{3}{5}\) - x = \(\dfrac{1}{10}\)
d) x . \(\dfrac{3}{4}\) = \(\dfrac{9}{20}\)
e) x : \(\dfrac{1}{7}\) = 14
f) ( \(\dfrac{1}{4}\) + x ) . \(\dfrac{1}{2}\) = \(\dfrac{2}{5}\)
g) x . \(\dfrac{2}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{9}{12}\)
h) \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
k) \(3\dfrac{4}{5}\) - x = \(\dfrac{18}{5}\)
l) x . \(2\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
m) x . \(\dfrac{6}{11}\) + x . \(\dfrac{5}{11}\) = 2025
n) x . \(\dfrac{14}{9}\) - x . \(\dfrac{7}{9}\) + x . \(\dfrac{5}{9}\) = 2
Các bạn làm theo cách bình thường ở lớp 5 cho mính nhé!
Chú ý: dấu "." là dấu nhân.
a: x+2/5=1/2
=>x=1/2-2/5=5/10-4/10=1/10
b; x-2/5=2/7
=>x=2/7+2/5=10/35+14/35=24/35
c: 3/5-x=1/10
=>x=3/5-1/10=6/10-1/10=5/10=1/2
d: x*3/4=9/20
=>x=9/20:3/4=9/20*4/3=36/60=3/5
e: x:1/7=14
=>x=14*1/7=2
f: =>x+1/4=2/5:1/2=4/5
=>x=4/5-1/4=16/20-5/20=11/20
g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12
=>x=17/12:2/3=17/12*3/2=51/24=17/8
Tìm x, biết:
a) \(\dfrac{1}{20}\) - (x - \(\dfrac{8}{5}\)) = \(\dfrac{1}{10}\)
b) \(\dfrac{7}{4}\) - (x + \(\dfrac{5}{3}\)) = \(\dfrac{-12}{5}\)
c) x - [\(\dfrac{17}{2}\) - \(\left(\dfrac{-3}{7}+\dfrac{5}{3}\right)\)] = \(\dfrac{-1}{3}\)
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
\(\dfrac{3}{4}.\left(\dfrac{2}{5}-2x\right)-\dfrac{3}{5}.\left(\dfrac{1}{2}x-\dfrac{5}{4}\right)=\dfrac{9}{20}\)
Đề yêu cầu gì vậy em?
\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
a \(\sqrt{4x-20}+\sqrt{x-5}=4+3\sqrt{\dfrac{x-5}{9}}\)
b \(\sqrt{4x-20}+\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{4x-45}=4\)
Lời giải:
a. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}=4+3.\sqrt{\frac{1}{9}}.\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
b. Sửa đoạn 4x-45 thành 4x-20.
ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{4}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$
$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=\frac{12}{5}$
$\Leftrightarrow x-5=\frac{144}{25}=5,76$
$\Leftrightarrow x=10,76$ (tm)