2^5+12(x^3-20)=116
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`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2}{3}-\dfrac{5}{6}+\dfrac{1}{12}\)
`=`\(\dfrac{8}{12}-\dfrac{10}{12}+\dfrac{1}{12}\)
`=`\(-\dfrac{1}{12}\)
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\(\dfrac{3}{4}+\dfrac{3}{16}-\dfrac{1}{2}?\)
`=`\(\dfrac{12}{16}+\dfrac{3}{16}-\dfrac{8}{16}\)
`=`\(\dfrac{7}{16}\)
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\(\dfrac{2}{5}-\dfrac{4}{7}+\dfrac{1}{2}\)
`=`\(\dfrac{2}{5}+\dfrac{1}{2}-\dfrac{4}{7}\)
`=`\(\dfrac{4}{10}+\dfrac{5}{10}-\dfrac{4}{7}\)
`=`\(\dfrac{9}{10}-\dfrac{4}{7}\)
`=`\(\dfrac{63}{70}-\dfrac{40}{70}=\dfrac{23}{70}\)
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a) Ta có: \(-3x=7y=21z\)
\(\Rightarrow-3x\cdot\frac{1}{21}=7y\cdot\frac{1}{21}=21z\cdot\frac{1}{21}\)
\(\Rightarrow\frac{x}{-7}=\frac{y}{3}=\frac{z}{1}=\frac{5x}{-35}=\frac{10y}{30}=\frac{6z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{5x}{-35}=\frac{10y}{30}=\frac{6z}{6}=\frac{5x+10y+6z}{-35+30+6}=\frac{4}{1}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{5x}{-35}=4\rightarrow5x=-140\rightarrow x=-28\\\frac{10y}{30}=4\rightarrow10y=120\rightarrow y=12\\\frac{6z}{6}=4\rightarrow z=4\end{cases}}\)
Vậy x= -28; y=12; z=4
b) Ta có: \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{5}\rightarrow\frac{x}{6}=\frac{y}{15}\\\frac{y}{3}=\frac{z}{20}\rightarrow\frac{y}{15}=\frac{z}{100}\end{cases}}\)
\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{100}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{100}=k\)
\(\Rightarrow x=6k;y=15k;z=100k\)
\(y\cdot z=900\rightarrow15k\cdot100k=900\)
\(\rightarrow1500\cdot k^2=900\)
\(\rightarrow k^2=\frac{3}{5}\rightarrow k\varepsilon\varnothing\)
Vậy x;y;z ko có giá trị thỏa mãn
c) Ta có: \(\frac{x}{2}=\frac{y}{5}=\frac{x^2}{4}=\frac{y}{25}^2\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{4}=\frac{y^2}{25}=\frac{x^2+y^2}{4+25}=\frac{116}{29}=4\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{4}=4\rightarrow x^2=16\rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\\\frac{y^2}{25}=4\rightarrow y^2=100\rightarrow\orbr{\begin{cases}y=10\\y=-10\end{cases}}\end{cases}}\)\(\Rightarrow\frac{x^2}{4}=4\rightarrow x^2=16\rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
\(\frac{y^2}{25}=4\rightarrow y^2=100\rightarrow\orbr{\begin{cases}y=10\\y=-10\end{cases}}\)
Vậy (x;y) = (4;10); (-4;-10)
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a) \(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2+x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\end{cases}}}\Rightarrow\)Vô lí
b)\(\Leftrightarrow\left(\frac{x+106}{3}-2\right)+\left(\frac{x+116}{4}-4\right)+\left(\frac{x+130}{5}-6\right)+\left(\frac{x-148}{6}-8\right)=0\Leftrightarrow\frac{x+100}{3}+\frac{x+100}{4}+\frac{x+100}{5}+\frac{x+100}{6}=0\Leftrightarrow\left(x+100\right)\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)=0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
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\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)
\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)
hay x=1
b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)
c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)
d: \(\Leftrightarrow-8< x< 3\)
hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b. \(\dfrac{x+106}{3}+\dfrac{x+116}{4}+\dfrac{x+130}{5}+\dfrac{x+148}{6}=0\)\(\Leftrightarrow\dfrac{x+106}{3}+\dfrac{x+116}{4}+\dfrac{x+130}{5}+\dfrac{x+148}{6}-20=0\)\(\Leftrightarrow\dfrac{x+106}{3}-2+\dfrac{x+116}{4}-4+\dfrac{x+130}{5}-6+\dfrac{x+148}{6}-8=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}\ne0\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy PT có nghiệm \(x=-100\)
\(x^4+x^3+2x^2+x+1=0\\ \Leftrightarrow\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)=0\\ \Leftrightarrow x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x^2+x+1\right)\left(x^2+1\right)=0\\ \)
Vì x^2+x+1\(>0\) với mọi x và x^2+1\(>0\) với mọi x nên (x^2+x+1)(x^2+1)>0 với mọi x
Vậy phương trình vô nghiệm
Ta có : 2 ^ 5 + 12 . ( x ^ 3 - 20 ) = 116
=> 32 + 12 . ( x ^ 3 - 20 ) = 116
=> 12 . ( x ^ 3 - 20 ) = 116 - 32
=> 12 . ( x ^ 3 - 20 ) = 84
=> x ^ 3 - 20 = 84 : 12
=> x ^ 3 - 20 = 7
=> x ^ 3 = 7 + 20
=> x ^ 3 = 27
=> x ^ 3 = 3 ^ 3
=> x = 3
Vậy x = 3
\(2^5+12\left(x^3-20\right)=116\)
\(\Rightarrow32+12\left(x^3-20\right)=116\)
\(\Rightarrow12\left(x^3-20\right)=116-32=84\)
\(\Rightarrow x^3-20=84:12=7\)
\(\Rightarrow x^3=7+20=27\)
\(\Rightarrow x^3=3^3\Rightarrow x=3\)
Vậy \(x=3\)
~ Học tốt ~