\(\text{|}\dfrac{7}{5}x+\dfrac{2}{3}\text{|}\text{=}\text{|}\dfrac{4}{3}x-\dfrac{1}{4}\text{|}\)
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\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
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\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
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`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
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d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
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`|5/4 x-7/2| -|5/8 x +3/5|=0`
`|5/4 x-7/2|=|5/8 x+3/5|`
\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
a) Thay x + 3y - 2z vào biểu thức ta có:
\(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhua ta có:
\(\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = \dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\)
=\(\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)
=\(\dfrac{36 + 9}{9}\) = 5
=> \(\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6
=>
=>
Vậy ...
(Bạn dựa theo cách này và lm những bài tiếp nhé!)
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c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
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\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)
\(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)
Bảng xét dấu:
\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)
\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)
Đặt \(f\left(x\right)=x^2-3x-7.\)
\(f\left(x\right)=x^2-3x-7.\)
\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
`|7/5 x+2/3| = |4/3 x-1/4|`
\(\left[{}\begin{matrix}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=-\dfrac{55}{4}\\x=-\dfrac{25}{164}\end{matrix}\right.\)