1,(x+7).x-2)>0
2.\(\dfrac{37-x}{x+13}\)=\(\dfrac{3}{7}\)
3.\(^{\dfrac{x-3}{x+8}}\)<0
Khó qua ai giúp vs.Ai trả lời nhanh cho một like
B1 : Tìm x,y
a) \(\dfrac{x}{-15}=\dfrac{60}{x}\)
b)\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)
c)\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
d) \(\dfrac{x}{2}=\dfrac{y}{5}=xy=10\)
Giúp tui đi :< Tui tick
d: Đặt x/2=y/5=k
=>x=2k; y=5k
Ta có: xy=10
nên k2=1
Trường hợp 1: k=1
=>x=2; y=5
Trường hợp 2: k=-1
=>x=-2; y=-5
a)\(\dfrac{2}{3}\)x-1=\(\dfrac{3}{2}\)
b)| 5x - \(\dfrac{1}{2}\)| - \(\dfrac{2}{7}\)= 25%
c)\(\dfrac{x-3}{4}\)=\(\dfrac{16}{x-3}\)
d)\(\dfrac{-8}{13}\)+\(\dfrac{7}{17}+\dfrac{21}{31}\)<x≤\(\dfrac{-9}{14}\)+4+\(\dfrac{5}{-14}\)(xϵZ)
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
h) \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
i) \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)
k) \(\dfrac{13}{x-1}+\dfrac{5}{2x-2}-\dfrac{6}{3x-3}\)
`h)x/2-1/x=1/12(x ne 0)`
`<=>6x^2-12=x`
`<=>6x^2-x-12=0`
`<=>6x^2-9x+8x-12=0`
`<=>3x(2x-3)+4(2x-3)=0`
`<=>(2x-3)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\)
`i)x^2-7/6x+1/3=0`
`<=>6x^2-7x+2=0`
`<=>6x^2-3x-4x+2=0`
`<=>3x(2x-1)-2(2x-1)=0`
`<=>(2x-1)(3x-2)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\)
Câu cuối không có dấu "=" nên không tìm được x :v
- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v
h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)
\(\Leftrightarrow12\left(x^2-2\right)-2x=0\)
\(\Leftrightarrow12x^2-2x-24=0\)
\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2+34}{12}=\dfrac{36}{12}=3\\x_2=\dfrac{2-34}{12}=\dfrac{-32}{12}=-\dfrac{8}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{3;-\dfrac{8}{3}\right\}\)
i) Ta có: \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
tìm x, biết:
e) \(\dfrac{2}{7-x}\)=\(\dfrac{18}{45}\) f)\(\dfrac{2x+3}{3}\)=\(\dfrac{50}{15}\) g)\(2\dfrac{x}{7}\)=\(\dfrac{75}{35}\) h)\(2\dfrac{3}{x}\)=\(\dfrac{13}{x}\)(x khác 0)
e: =>2/7-x=2/5
=>7-x=5
=>x=2
f: =>2x+3/3=10/3
=>2x+3=10
=>2x=7
=>x=7/2
g: =>(14+x)/7=15/7
=>x+14=15
=>x=1
h: =>(2x+3)/x=13/x
=>2x+3=13
=>2x=10
=>x=5
1.\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(2,\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
\(3,\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\)
\(1\text{)}\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
\(\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow-x-13=21x+273\)
\(\Leftrightarrow-22x=286\)
\(\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}\)
\(3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6\)
\(\Leftrightarrow x-2=7x-6\)
\(\Leftrightarrow-6x=-4\)
\(\Rightarrow x=\dfrac{2}{3}\)
1.(x−5)2+3(x−5)=0
=>(x-5)(x-5)+3.(x-5)=0
=>(x-5).(x-5+3)=0
=>x-5=0 hoặc x-2=0
=>x=5 hoặc x=2
2)\(\dfrac{2x-1}{3}\)-\(\dfrac{5x+2}{7}\)=x+13
=>\(\dfrac{7.\left(2x-1\right)}{7.3}\)-\(\dfrac{3.\left(5x+2\right)}{3.7}\)=\(\dfrac{21.\left(x+13\right)}{21}\)
=>khử mẫu:
=>7.(2x-1)-3.(5x+2)=21.(x+13)
=>14x-7-15x-6=21x+273
=>14-7-15x-6-21x-273=0
=>-36x-272=0
=>-36x=272
=>x=...
Tìm x, biết:
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\) b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\) c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\) d) \(\dfrac{2}{3}.x-\dfrac{7}{6}=\dfrac{5}{2}\)
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
\(a,\)\(x:\dfrac{6}{13}=\dfrac{13}{7}\)
\(x=\dfrac{13}{7}.\dfrac{6}{13}\)
\(x=\dfrac{6}{7}\)
b,\(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
\(x=\dfrac{13}{15}.\dfrac{7}{4}\)
\(x=\dfrac{91}{60}\)
a: Ta có: \(x:\dfrac{6}{13}=\dfrac{13}{7}\)
\(\Leftrightarrow x=\dfrac{13}{7}\cdot\dfrac{6}{13}\)
hay \(x=\dfrac{6}{7}\)
b: Ta có: \(\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{13}{15}\)
hay \(x=\dfrac{91}{60}\)
g) \(3-\dfrac{2}{2x-3}=\dfrac{2}{5}=\dfrac{2}{9-6x}-\dfrac{3}{2}\)
h) \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
i) \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)
k) \(\dfrac{13}{x-1}+\dfrac{5}{2x-2}-\dfrac{6}{3x-3}\)
m) \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)
n) \(\left(\dfrac{3}{2}-\dfrac{5}{11}-\dfrac{3}{13}\right)\left(2x-2\right)=\left(-\dfrac{3}{4}+\dfrac{5}{22}+\dfrac{3}{26}\right)\)
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)
\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)
\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)
Vậy ...
i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)
\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)
Vậy ...
h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)
\(\Leftrightarrow12x^2-24-2x=0\)
\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)
m) Ta có: \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{19}{10}:x=2\)
hay \(x=\dfrac{19}{20}\)
Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)
a)x=\(\dfrac{1}{4}+\dfrac{2}{13}\)
b)\(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
c)\(\dfrac{-8}{3}+\dfrac{-1}{3}\le x\le\dfrac{-2}{7}+\dfrac{-5}{7}\)
d)\(\dfrac{-5}{6}+\dfrac{8}{3}\dfrac{29}{-x}\le x\le\dfrac{-1}{2}+2+\dfrac{5}{2}\)