4-x=22/3
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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}.
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
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: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
36x4=144
50x3=150
22x6=132
35x4=140
46:2=23
96:3=32
60:3=20
88:4=22
\(x\times\dfrac{2}{3}+x\times\dfrac{4}{3}=\dfrac{22}{5}\\ x\times\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{22}{3}\\ x\times2=\dfrac{22}{3}\\ x=\dfrac{11}{3}\)
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\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\cdot\left(x+3\right)}=\frac{22}{35}\)
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{x+3}=\frac{1}{35}\)
\(\Rightarrow x+3=35\)
\(\Rightarrow x=32\)
\(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\left(x+3\right)}=\frac{22}{35}\)
\(\frac{3}{4}\left(\frac{4}{2\cdot5}+\frac{4}{5\cdot8}+...+\frac{4}{x\left(x+3\right)}\right)=\frac{3}{4}\cdot\frac{22}{35}\)
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{x\left(x+3\right)}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{2}-\frac{1}{x+3}=\frac{33}{70}\)
\(\frac{1}{x+3}=\frac{1}{35}\)
\(\Rightarrow x+3=35\)
\(\Rightarrow x=32\)
Vậy x=32
\(\dfrac{93}{17}:x+\left[\dfrac{-4}{17}\right]:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)
\(=>\left(\dfrac{93}{17}+\dfrac{-4}{17}\right):x+\dfrac{11}{26}=\dfrac{4}{11}\)
\(=>\dfrac{89}{17}:x+\dfrac{11}{26}=\dfrac{4}{11}\)
\(=>\dfrac{89}{17}:x=\dfrac{4}{11}-\dfrac{11}{26}\)
\(=>\dfrac{89}{17}:x=\dfrac{-17}{286}\)
\(=>x=\dfrac{89}{17}:\dfrac{-17}{286}\)
\(=>x=\dfrac{-25454}{289}=-88\dfrac{22}{289}\)
\(3\left(x-4\right)-\left(8-x\right)=22\)
\(\Leftrightarrow3x+x-12-8=22\)
\(\Leftrightarrow4x=42\)
\(\Leftrightarrow x=10,5\)
<=> x=\(4-\dfrac{22}{3}\)
<=> x=\(\dfrac{12-22}{3}\)
<=> x=\(\dfrac{-10}{3}\)
x = 22/3-4
x = 10/3.