\(\frac{11}{12}\)-(\(\frac{2}{5}\)+x)=\(\frac{2}{3}\)*(6x+1)
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\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\cdot\left(6x+1\right)\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}.\left(6x+1\right)\)
\(\Rightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}.6x+\frac{2}{3}\)
\(\Rightarrow\frac{55-24}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-\frac{2}{3}=4x+x=5x\)
\(\Rightarrow5x=-\frac{11}{60}\)
\(\Rightarrow x=\frac{-11}{300}\)
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Bài 8: Tìm số nguyên x biết
a) \(\left(\frac{-12}{27}+\frac{2}{3}\right)+\frac{-2}{9}\le x\le\left(\frac{11}{7}+\frac{2}{5}\right)+\frac{7}{5}+\frac{3}{7}\) b\(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
c)\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +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
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Giải phương trình sau:
a) \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
b) \(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)
c) \(\frac{x-2}{x+2}+\frac{3}{2-x}=\frac{2\left(x-11\right)}{x^2-4}\)
d) \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)
a) ĐKXĐ: $x\neq 1$
PT \(\Leftrightarrow \frac{x^2+x+1+2(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow \frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)
\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow 2x^2-3x+1=0\)
\(\Leftrightarrow (x-1)(2x-1)=0\)
Mà $x\neq 1$ nên $2x-1=0\Rightarrow x=\frac{1}{2}$ là nghiệm
b) ĐK: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3-x}{2-x}=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{3-x}{2-x}=\frac{6-x}{3(x^2-4)}\)
\(\Leftrightarrow \frac{1}{x+2}+\frac{3-x}{x-2}=\frac{6-x}{3(x-2)(x+2)}\)
\(\Leftrightarrow \frac{-x^2+2x+4}{(x-2)(x+2)}=\frac{6-x}{3(x-2)(x+2)}\)
\(\Rightarrow 3(-x^2+2x+4)=6-x\)
\(\Leftrightarrow -3x^2+7x+6=0\)
\(\Leftrightarrow (x-3)(3x+2)=0\Rightarrow x=3\) hoặc $x=-\frac{2}{3}$
Vậy........
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c) ĐK: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2(x-11)}{x^2-4}\)
\(\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{x^2-7x-2}{(x-2)(x+2)}=\frac{2x-22}{(x-2)(x+2)}\)
\(\Rightarrow x^2-7x-2=2x-22\)
\(\Leftrightarrow x^2-9x+20=0\Leftrightarrow (x-4)(x-5)=0\Rightarrow x=4\) hoặc $x=5$
(đều thỏa mãn)
d) ĐK: \(x^2-6x+7\neq 0\)
PT \(\Leftrightarrow (x^2-6x+7)+\frac{14}{x^2-6x+7}-9=0\)
\(\Rightarrow (x^2-6x+7)^2-9(x^2-6x+7)+14=0\)
\(\Leftrightarrow (x^2-6x+7-2)(x^2-6x+7-7)=0\)
\(\Leftrightarrow (x^2-6x+5)(x^2-6x)=0\)
\(\Leftrightarrow (x-1)(x-5)x(x-6)=0\)
\(\Rightarrow x\in \left\{1;5;0;6\right\}\) (đều thỏa mãn)
Vậy.........
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Tim so huu ti x
a)\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\))
b)\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
c)\(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{2}{9-6x}-\frac{3}{2}\)
e)\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
\(ĐKXĐ:x\ne3;x\ne5;x\ne4;x\ne6\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Rightarrow\frac{x}{x-3}-\frac{x}{x-5}-\frac{x}{x-4}+\frac{x}{x-6}=0\)
\(\Rightarrow x\left(\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}=0\left(1\right)\end{cases}}\)
\(\left(1\right)\Rightarrow\frac{1}{x-3}+\frac{1}{x-6}=\frac{1}{x-5}+\frac{1}{x-4}\)
\(\Rightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-5\right)\left(x-4\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\left(tm\right)\\\left(x-3\right)\left(x-6\right)=\left(x-5\right)\left(x-4\right)\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow x^2-9x+18=x^2-9x+20\)
\(\Leftrightarrow0=2\left(L\right)\)
Vậy pt có 2 nghiệm \(\left\{0;\frac{9}{2}\right\}\)
Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b0
1. a, frac{5x-2}{3}frac{5-3x}{2}; b, frac{10x+3}{12}1+frac{6+8x}{9}
c, 2left(x+frac{3}{5}right)5-left(frac{13}{5}+xright); d, frac{7}{8}x-5left(x-9right)frac{20x+1,5}{6}
e, frac{7x-1}{6}+2xfrac{16-x}{5}; f, 4 (0,5-1,5x)frac{5x-6}{3}
g, frac{3x+2}{2}-frac{3x+1}{6}frac{5}{3}+2x; h, frac{x+4}{5}.x+4frac{x}{3}-frac{x-2}{2}
i, frac{4x+3}{5}-frac{6x-2}{7}fr...
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Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Đây là những bài cơ bản mà bạn!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
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Tìm x biết :
\(\left[\frac{6:\frac{3}{5}-1\frac{1}{16}.\frac{6}{7}}{4\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}}-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{15}\right).\frac{12}{49}}{3\frac{1}{3}+\frac{2}{9}}\right].x=2\frac{23}{96}\)
\(x=\frac{903}{391}\)
Bài này sử dụng MTCT đó bạn!
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a)tìm x,y biết:\(\frac{x}{12}=\frac{y}{3}vàx-y=36\)
b)tìm x biết:\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)
=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)
\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(48;12\right).\)
b)
\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
⇒ \(\frac{5}{3}x=\frac{1}{21}\)
⇒ \(x=\frac{1}{21}:\frac{5}{3}\)
⇒ \(x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}.\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
⇒ \(x-\frac{1}{2}=\frac{1}{3}\)
⇒ \(x=\frac{1}{3}+\frac{1}{2}\)
⇒ \(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}.\)
Có 1 câu bạn đăng mình làm ở dưới rồi mà.
Chúc bạn học tốt!
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a)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\)x/12=4 suy ra x=12.4=48
y/3=4 suy ra y=3.4 =12
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=\frac{-3}{20}\)
\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=2\)
suy ra x-2/5=2 hoac x-2/5=-2
\(x-\frac{2}{5}=2\)
\(x=\frac{12}{5}\)
\(x-\frac{2}{5}=-2\)
\(x=\frac{-8}{5}\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
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a)tìm x,y biết:
Ta có : \(\dfrac{x}{12} = \dfrac{y}{3}\) \(\Rightarrow\) \(\dfrac{x-y}{12-3} = \dfrac{36}{9} = 4\)
b)tìm x biết:
23+53x=57
\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{5}{7} - \dfrac{2}{3}\)
\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{1}{21}\)
\(\Leftrightarrow\) \(x = \dfrac{1}{35}\)
1112−(25+x)=23
\(\Leftrightarrow\) \(\dfrac{2}{5} +x\) = \(\dfrac{2}{3} - \dfrac{11}{12}\)
\(\Leftrightarrow\) \(\dfrac{2}{5} + x = \dfrac{-1}{4}\)
\(\Leftrightarrow\) \(x = \dfrac{-13}{20}\)
|x−25|+34=114
\(\Leftrightarrow\) \(x - \dfrac{2}{5} = 2\)
\(\Leftrightarrow\) \(x = \dfrac{8}{5}\)
\(\Leftrightarrow\) \(x - \dfrac{1}{2} = \dfrac{1}{19683}\)
\(\Leftrightarrow\) \(x = \dfrac{19685}{39366}\)
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