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a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{x+y+z}{2+3+5}=\frac{90}{10}=9\)
\(\Leftrightarrow\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.5=45\end{cases}}\)
b) \(2x=3y\Leftrightarrow\frac{x}{15}=\frac{y}{10},2y=5z\Leftrightarrow\frac{y}{10}=\frac{z}{4}\)
suy ra \(\frac{x}{15}=\frac{y}{10}=\frac{z}{4}\).
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{15}=\frac{y}{10}=\frac{z}{4}=\frac{x-z}{15-4}=\frac{11}{11}=1\)
\(\Leftrightarrow\hept{\begin{cases}x=15.1=15\\y=10.1=10\\z=4.1=4\end{cases}}\)
c) \(\frac{x}{y}=\frac{3}{4}\Leftrightarrow\frac{x}{9}=\frac{y}{12},\frac{y}{z}=\frac{3}{5}\Leftrightarrow\frac{y}{12}=\frac{z}{20}\)
suy ra \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{2.9-3.12+20}=\frac{6}{2}=3\)
\(\Leftrightarrow\hept{\begin{cases}x=3.9=27\\y=3.12=36\\z=3.20=60\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5,(9) = 5 + 0,(9) = 5 + 9.0,(1) \(=5+9.\frac{1}{9}\)= 5 + 1 = 6
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`7,`
`a, B+A=4x-2x^2+3`
`-> B=(4x-2x^2+3)-A`
`-> B=(4x-2x^2+3)-(x^2-2x+1)`
`B=4x-2x^2+3-x^2+2x-1`
`B=(-2x^2-x^2)+(4x+2x)+(3-1)`
`B=-3x^2+6x+2`
`b, C-A=-x+7`
`-> C=(-x+7)+A`
`-> C=(-x+7)+(x^2-2x+1)`
`-> C=-x+7+x^2-2x+1`
`C=x^2+(-x-2x)+(7+1)`
`C=x^2-3x+8`
`c,`
`A-D=x^2-2`
`-> D= A- (x^2-2)`
`-> D=(x^2-2x+1)-(x^2-2)`
`D=x^2-2x+1-x^2+2`
`D=(x^2-x^2)-2x+(1+2)`
`D=-2x+3`
`6,`
`a,`
`P+Q=4x-2x^2+3`
`-> Q=(4x-2x^2+3)-P`
`-> Q=(4x-2x^2+3)-(3x^2+x-1)`
`Q=4x-2x^2+3-3x^2-x+1`
`Q=(-2x^2-3x^2)+(4x-x)+(3+1)`
`Q=x^2+3x+4`
`b,`
`x^2-5x+2-P=H`
`-> H= (x^2-5x+2)-(3x^2+x-1)`
`H=x^2-5x+2-3x^2-x+1`
`H=(x^2-3x^2)+(-5x-x)+(2+1)`
`H=-4x^2-6x+3`
`c,`
`P-R=5x^2-3x-4`
`-> R= P- (5x^2-3x-4)`
`-> R=(3x^2+x-1)-(5x^2-3x-4)`
`R=3x^2+x-1-5x^2+3x+4`
`R=(3x^2-5x^2)+(x+3x)+(-1+4)`
`R=-2x^2+4x+3`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 1
1.\(x\left(x+3\right)\)
\(=x^2+3x\)
2.\(3x\left(x+2\right)\)
\(=3x^2+6x\)
3,\(x^2\left(3x-1\right)\)
\(=3x^3-x^2\)
4.\(-5x^3\left(3x^2-7\right)\)
\(=-15x^5+35x^3\)
5.\(3x\left(5x^2-2x-1\right)\)
\(=15x^3-6x^2-3x\)
6.\(-x^2\left(5x^3-x-\dfrac{1}{2}\right)\)
\(=-5x^5+x^3+\dfrac{x^2}{2}\)
7.\(\left(x^2+2x-3\right).\left(-x\right)\)
\(=-x^3-2x^2+3x\)
8.\(4x^3\left(-2x^2+4x^4-3\right)\)
\(=-8x^5+16x^7-12x^3\)
9.\(-5x^2\left(3x^2-2x+1\right)\)
\(=-15x^4+10x^3-5x^2\)
10.\(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+16x^7-28x^6+12x^5\)
11.\(\left(x+2\right)\left(x+3\right)\)
\(=x^2+3x+2x+6\)
12.\(\left(x-7\right)\left(x-5\right)\)
\(=x^2-5x-7x+35\)
13.\(\left(3x+5\right)\left(2x-7\right)\)
\(=6x^2-21x+10x-35\)
14.\(\left(x-3\right)\left(x^2-2x-1\right)\)
\(x^3-2x^2-x-3x^2+6x+3\)
15.\(\left(2x-1\right)\left(x^2-5x+3\right)\)
\(=2x^3-10x^2+6x-x^2+5x-3\)
16.\(\left(x-5\right)\left(-x^2+x-1\right)\)
\(=-x^3+x^2-x+5x^2-5x+5\)
17,\(\left(\dfrac{1}{2}x+3\right)\left(2x^2-4x-6\right)\)
\(=x^3-2x^2-3x+6x^2-12x-18\)
P/s:mình làm hơi tắt tại bài dài quá:))
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 5:
\(\dfrac{x}{y}=a\Rightarrow\dfrac{x}{a}=\dfrac{y}{1}=\dfrac{x-y}{a-1}=\dfrac{x+y}{a+1}\)
\(\Rightarrow\dfrac{x+y}{x-y}=\dfrac{a+1}{a-1}\)
Câu 6:
\(9x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{3x}{15}=\dfrac{2y}{18}=\dfrac{3x-2y}{15-18}=\dfrac{12}{-3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).5=-20\\y=\left(-4\right).9=-36\end{matrix}\right.\)
Câu 7:
\(\dfrac{x}{-5}=\dfrac{y}{7}=\dfrac{x+y}{-5+7}=\dfrac{-10}{2}=-5\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-5\right).\left(-5\right)=25\\y=\left(-5\right).7=-35\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{20}+19+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(S=19\)
Nếu ko hiểu kb vs mền rồi mền giải thích cho