NA
5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
4 tháng 6 2016
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
30 tháng 5 2020
1) Ta có: \(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|-\frac{3}{2}=\frac{1}{4}\)
⇔\(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
⇔\(\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{7}{4}:2=\frac{7}{4}\cdot\frac{1}{2}=\frac{7}{8}\)
⇔\(\left[{}\begin{matrix}\frac{1}{2}x-\frac{3}{8}=\frac{7}{8}\\\frac{1}{2}x-\frac{3}{8}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{10}{8}\\\frac{1}{2}x=\frac{-4}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{10}{8}:\frac{1}{2}=\frac{10}{8}\cdot2=\frac{20}{8}=\frac{5}{2}\\x=\frac{-4}{8}:\frac{1}{2}=-\frac{4}{8}\cdot2=-\frac{8}{8}=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{2};-1\right\}\)
2) Ta có: \(-5\cdot\left(x+\frac{1}{5}\right)-\frac{1}{2}\cdot\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
⇔\(-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{3}{2}x+\frac{5}{6}=0\)
\(\Leftrightarrow-7x+\frac{1}{6}=0\)
\(\Leftrightarrow-7x=-\frac{1}{6}\)
hay \(x=\frac{1}{42}\)
Vậy: \(x=\frac{1}{42}\)
3) Ta có: \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-\frac{3}{2}-5x-3+x-\frac{1}{5}=0\)
\(\Leftrightarrow-x-\frac{47}{10}=0\)
⇔\(-x=\frac{47}{10}\)
hay \(x=\frac{-47}{10}\)
Vậy: \(x=\frac{-47}{10}\)
4) Ta có: \(\frac{3}{4}-2\left|2x-0,125\right|=2\)
\(\Leftrightarrow2\left|2x-\frac{1}{8}\right|=\frac{3}{4}-2=-\frac{5}{4}\)
⇔\(\left|2x-\frac{1}{8}\right|=-\frac{5}{8}\)(vô lý)
Vậy: x∈∅
5) Ta có: \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
⇔\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\\\frac{1}{2}x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{24}:\frac{1}{2}=\frac{29}{24}\cdot2=\frac{29}{12}\\x=-\frac{13}{24}:\frac{1}{2}=-\frac{13}{24}\cdot2=-\frac{13}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
S
subjects
VIP
9 tháng 3
\(a;x:\left(9\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{\dfrac{2}{5}+\dfrac{4}{9}-\dfrac{5}{11}}{\dfrac{8}{5}+\dfrac{16}{9}-\dfrac{20}{11}}\\ x:8=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}\cdot8=2\)
\(b;\left|2x-\dfrac{1}{3}\right|-\left(-2^2\right)=4\left(\dfrac{1}{-2}\right)^3\\ \left|2x-\dfrac{1}{3}\right|+4=-\dfrac{1}{2}\\ \left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{2}-4=-\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{9}{2}\Rightarrow x=-\dfrac{25}{12}\\2x-\dfrac{1}{3}=\dfrac{9}{2}\Rightarrow x=\dfrac{29}{12}\end{matrix}\right.\)
S
subjects
VIP
9 tháng 3
\(a;\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}=\dfrac{20}{27}\\ b;\left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(-\dfrac{10}{11}\right)\\ \dfrac{9}{11}-x=-\dfrac{2}{11}\\ x=\dfrac{9}{11}-\left(-\dfrac{2}{11}\right)=\dfrac{9}{11}+\dfrac{2}{11}\\ x=1\\ c;-\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\\ -\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\\ -\dfrac{11}{12}x=-\dfrac{11}{12}\\ x=\left(-\dfrac{11}{12}\right):\left(-\dfrac{11}{12}\right)=1\)
\(d;-\dfrac{5}{4}-\left(1\dfrac{1}{2}+x\right)=4,5\\ \Leftrightarrow-\dfrac{5}{4}-\left(\dfrac{3}{2}+x\right)=4,5\\\dfrac{3}{2}+x=-\dfrac{5}{4}-4,5\\ \dfrac{3}{2}+x=-\dfrac{23}{4}\\ x=-\dfrac{23}{4}-\dfrac{3}{2}\\ x=-\dfrac{29}{4}\\ đ;\left(\dfrac{3}{4}-x:\dfrac{2}{15}\right)\cdot\dfrac{1}{5}=-2,6\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-2,6:\dfrac{1}{5}\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-13\\ x:\dfrac{2}{15}=\dfrac{3}{4}-\left(-13\right)\\ x:\dfrac{2}{15}=\dfrac{55}{4}\\ x=\dfrac{55}{4}\cdot\dfrac{2}{15}=\dfrac{11}{6}\\ e;3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}=-\dfrac{10}{3}\)
\(f;\left(1-2x\right)\cdot\dfrac{4}{5}=\left(-2\right)^3\\ \left(1-2x\right)\cdot\dfrac{4}{5}=-8\\ 1-2x=-8:\dfrac{4}{5}=-10\\ 2x=1-\left(-10\right)=11\\ x=\dfrac{11}{2}\\ g;\dfrac{1}{6}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{24}\Rightarrow x=\dfrac{3}{4}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{24}\Rightarrow x=\dfrac{7}{12}\end{matrix}\right.\)
(x-5/8)^2= 1/4-(-2)
(x-5/8)^2= 9/4
(x-5/8)^2= (3/2)^2
=> x-5/8= 3/2
x= 3/2+5/8
x= 17/8
chúc e học tốt
Ta có: \(\frac14-\left(x-\frac58\right)^2=-2\)
=>\(\left(x-\frac58\right)^2=\frac14+2=\frac94\)
=>\(\left[\begin{array}{l}x-\frac58=\frac32\\ x-\frac58=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac58+\frac32=\frac58+\frac{12}{8}=\frac{17}{8}\\ x=\frac58-\frac32=\frac58-\frac{12}{8}=-\frac78\end{array}\right.\)