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NT
1
NG
1
28 tháng 4 2019
\(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}+\frac{x}{13\cdot16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\cdot\frac{15}{16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}:\frac{15}{16}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}\cdot\frac{16}{15}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{1}\cdot\frac{8}{3}\)
\(\Rightarrow\frac{x}{3}=\frac{8}{3}\Leftrightarrow x=8\)
MV
2 tháng 5 2017
1.
\(\left(x+\dfrac{1}{3}\right)^2=16\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)
2.
\(x+\dfrac{3}{10}=\dfrac{7}{5}\\ x=\dfrac{7}{5}-\dfrac{3}{10}\\ x=\dfrac{11}{10}\)
2 tháng 5 2017
\(\left(x+\dfrac{1}{3}\right)^2=16\)
\(\left[{}\begin{matrix}\left(x+\dfrac{1}{3}\right)^2=4^2\\\left(x+\dfrac{1}{3}\right)^2=\left(-4\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4-\dfrac{1}{3}\\x=-4-\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)
HP
1
5 tháng 12 2021
\(\text{a)}3:x-10=-12\)
\(3:x\) \(=\left(-12\right)+10=-2\)
\(x\) \(=3:\left(-2\right)=-1,5\)
\(\text{b)}\left(2-x\right):\left(-4\right)-5=11\)
\(\left(2-x\right):\left(-4\right)\) \(=11+5=16\)
\(\left(2-x\right)\) \(=16x\left(-4\right)=-64\)
\(x\) \(=2-\left(-64\right)=66\)
\(\text{c)}x:\left(-5\right)+10=-60\)
\(x:\left(-5\right)\) \(=\left(-60\right)-10=-70\)
\(x\) \(=\left(-70\right)x\left(-5\right)=350\)
\(\left(3-2x\right)^2=16\)
\(\left(3-2x\right)^2=\pm\left(4\right)^2\)
\(\text{Vậy 3-2x=4 hoặc }\) \(3-2x=-4\)
\(2x=3-4=-1\) \(2x=3-\left(-4\right)=7\)
\(x=\left(-1\right):2=-0,5\) \(x=7:2=3,5\)
\(\text{Câu b với c mik lỡ làm ngược lại rồi bạn thông cảm}\)
\(\text{Hok tốt!}\)
\(\text{@Kaito Kid}\)
28 tháng 3 2020
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
`5/[x-2]=10/16`
`10(x-2)=16.5=80`
`x-2=80:10=8`
`x=8+2=10`
\(\dfrac{5}{x-2}=\dfrac{10}{16}\)
\(\Rightarrow5.16=\left(x-2\right).10\)
\(\Rightarrow\left(x-2\right).10=80\)
\(\Rightarrow x-2=80:10\)
\(\Rightarrow x-2=8\)
\(\Rightarrow x=8+2\)
\(\Rightarrow x=10\)