22|x-1|-3=5
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CM
24 tháng 6 2019
Ta có: (x + 2)2 – 4 ≥ (x + 3)(x + 5) – x
⇔ x2 + 4x + 4 – 4 ≥ x2 + 5x + 3x + 15 – x
⇔ –3x ≥ 15 ⇔ x ≤ –5
Tập nghiệm: S = {x | x ≤ –5}.
26 tháng 7 2023
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
TN
1
27 tháng 12 2021
\(\frac{3}{22}\times\frac{3}{11}\times22\)
1.\(=\frac{3\times3\times22}{22\times11\times1}\)
\(=\frac{3\times3}{11\times1}\)
\(=\frac{9}{11}\)
2. \(=\frac{9}{242}\times22\)
\(=\frac{198}{242}\)
\(=\frac{9\times2\times11}{2\times11\times11}\)
\(=\frac{9}{11}\)
\(\left(\frac{1}{2}+\frac{1}{3}\right)\times\frac{2}{5}\)
1.\(=\frac{5}{6}\times\frac{2}{5}\)
\(=\frac{5\times2}{3\times2\times5}\)
\(=\frac{1}{3}\)
2.\(\)\(=\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{2}{5}\)
\(=\frac{1}{5}+\frac{2}{15}\)
\(=\frac{3}{15}+\frac{2}{15}\)
\(=\frac{5}{15}=\frac{1}{3}\)
Have a good day
PV
1
26 tháng 2 2019
3/22 x 3/11 x 22=
c1:\(\frac{3\times3\times22}{22\times11}=\frac{9}{11}\) c2:\(\frac{9}{242}\times22=\frac{9}{11}\)
( 1/2 + 1/3 ) x 2/5=
c1:\(\frac{5}{6}\times\frac{2}{5}=\frac{1}{3}\) c2:\(\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{2}{5}=\frac{1}{5}+\frac{2}{15}=\frac{1}{3}\)
18 tháng 4 2022
a. \(x-\dfrac{3}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{3}{2}\)
\(x=\dfrac{14}{6}=\dfrac{7}{3}\)
TL
1
LD
27 tháng 3 2022
\(a,2\left(x-1\right)+3=x+2\)
\(\Leftrightarrow2x-2+3=x+2\)
\(\Leftrightarrow2x-x=2+2-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(b,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)
\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(5+x\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-5;2\right\}\)
2 tháng 7 2021
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
D
3
12 tháng 3 2015
1, =4/9 x (3/7 + 2 x 4/7) - 1/9 - 3/9
= 4/9 x 1 - 1/9 - 3/9 = 0
2, = ( 22/5 x 5/22 ) x 12
= 1 x 12 = 12
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)