2(x-1/2)+3 (x-1/3)= -22
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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:\(\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{2}{5}=\frac{1}{5}+\frac{2}{15}=\frac{1}{3}\)
\(\frac{3}{22}\times\frac{3}{11}\times22\)
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
\(a,2\left(x-1\right)+3=x+2\)
\(\Leftrightarrow2x-2+3=x+2\)
\(\Leftrightarrow2x-x=2+2-3\)
\(\Leftrightarrow x=1\)
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 \(S=\left\{-5;2\right\}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
<=> \(\left(2x+3\right)^2-4x^2+1=22\)
<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
<=> \(3\left(4x+3\right)=21\)
<=> \(4x+3=7\)
<=> \(4x=4\)
<=> \(x=1\)
......................?
mik ko biết
mong bn thông cảm
nha ................
a. \(x-\dfrac{3}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{3}{2}\)
\(x=\dfrac{14}{6}=\dfrac{7}{3}\)
a) Ta có:
\(\dfrac{x+1}{3}=\dfrac{y+2}{2}=\dfrac{z+3}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được
\(\dfrac{x+1}{3}=\dfrac{y+2}{2}=\dfrac{z+3}{1}\)
\(=\dfrac{x+1-y-2+z+3}{3-2+1}\)
\(=\dfrac{22+2}{2}\)
\(=\dfrac{24}{2}=12\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{3}=12\\\dfrac{y+2}{2}=12\\\dfrac{z+3}{1}=12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\y+2=24\\z+3=12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=36-1=35\\y=24-2=22\\z=12-3=9\end{matrix}\right.\)
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}.
\(2\left[x-\frac{1}{2}\right]+3\left[x-\frac{1}{3}\right]=-22\)
\(\Rightarrow2x-1+3x-1=-22\)
\(\Rightarrow2x+3x-1-1=-22\)
\(\Rightarrow5x=-22+1+1\)
\(\Rightarrow5x=-20\Leftrightarrow x=-4\)