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a) \(x+\left(-7\right)=-20\)
\(\Rightarrow x=-20+7\)
\(\Rightarrow x=-13\)
Vậy \(x=-13\)
b) \(8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
c) \(|x|-7=-6\)
\(\Rightarrow|x|=-6+7\)
\(\Rightarrow|x|=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy \(x\in\left\{1;-1\right\}\)
d) \(5^2.2^2-7.|x|=65\)
\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)
\(\Rightarrow10^2-7.|x|=65\)
\(\Rightarrow100-7.|x|=65\)
\(\Rightarrow7.|x|=35\)
\(\Rightarrow|x|=5\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy \(x\in\left\{5;-5\right\}\)
e) \(37-3.|x|=2^3-4\)
\(\Rightarrow37-3.|x|=8-4\)
\(\Rightarrow37-3.|x|=4\)
\(\Rightarrow3.|x|=33\)
\(\Rightarrow|x|=11\)
\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)
Vậy \(x\in\left\{11;-11\right\}\)
f) \(|x|+|-5|=|-37|\)
\(\Rightarrow|x|+5=37\)
\(\Rightarrow|x|=32\)
\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)
Vậy \(x\in\left\{32;-32\right\}\)
g)\(5.|x+9|=40\)
\(\Rightarrow|x+9|=8\)
\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)
Vậy \(x\in\left\{-1;-17\right\}\)
h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
Vậy \(-3\le x\le4\)
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,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)
Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9
\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)
b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)
<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)
<=>x=-43/30
Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại
a,A=1863/623:2/7
A=1863/178
b,
ta có:
1863/178.x+5/6=-3/4
1863/178.x=-19/12
=>x=-1691/11178
\(a,\left(x\cdot6\frac{2}{7}+\frac{3}{7}\right)\cdot2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x\cdot\frac{44}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(\frac{44x}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{44x+3}{7}\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{11\cdot\left(44x+3\right)}{5\cdot7}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{15}{35}=-2\)
\(\Rightarrow\frac{484x+33-15}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=\frac{-70}{35}\)
\(\Rightarrow484x+18=\left(-70\right)\)
\(\Rightarrow484x=\left(-70\right)-18\)
\(\Rightarrow484x=-88\)
\(\Rightarrow x=-\frac{88}{484}=-\frac{2}{11}\)
\(b,x\cdot3\frac{1}{4}+\left(-\frac{7}{6}\right)\cdot x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\cdot\frac{13}{4}+\left(-\frac{7}{6}\right)\cdot x-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x}{4}+\left(-\frac{7x}{6}\right)-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x\cdot3}{12}+\left(-\frac{7x\cdot2}{12}\right)-\frac{5\cdot4}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x}{12}+\left(-\frac{14x}{12}\right)-\frac{20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x-14x-20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{25x-20}{12}=\frac{5}{12}\)
\(\Rightarrow25x-20=5\)
\(\Rightarrow25x=20+5\)
\(\Rightarrow25x=25\)
\(\Rightarrow x=1\)
a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
\(\frac{4}{7}\times x=\frac{1}{5}+\frac{2}{3}\)
\(\frac{4}{7}x=\frac{13}{15}\)
\(\Rightarrow x=\frac{91}{60}\)
các bài còn lại tương tự nha
mấy cái này dễ mà toán tìm x này là cơ bản!!
67865785685685785785774677567568568
Ta có : \(\frac{x-5}{6}=\frac{3-x}{7}\)
=> [x-5] \(\times\) 7 = [3-x] \(\times\)6
=> 7x -35 = 18 - 6x
=> 13x - 35= 18
=> 13x = 53=> x = \(\frac{53}{13}\)
cho minh đúng cái ........................