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a) Ta có :
\(x - y = 5\)
\(\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{x-y}{2-3}=\frac{5}{-1}=-5\)
\(\Rightarrow\hept{\begin{cases}x=-5 . 2 = -10\\y=-5.3=-15\end{cases}}\)
b) Ta có :
\(x - y = 9\)
\(\frac{x}{-2}=\frac{y}{-5}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x}{-2}=\frac{y}{-5}=\frac{x-y}{-2-\left(-5\right)}=\frac{9}{3}=3\)
\(\Rightarrow\hept{\begin{cases}x=3. \left(-2 \right)= -6\\y=3 . \left(-5\right) = -15\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-0,75\right|-3\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=3\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{2}\\x-\dfrac{3}{4}=-\dfrac{7}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{4}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
\(PT\Rightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{7}{2}\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{2}\\x-\dfrac{3}{4}=-\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{4}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
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\(=\left[{}\begin{matrix}\dfrac{5}{8}-x+\dfrac{1}{4}-\dfrac{3}{2}\\x-\dfrac{5}{8}+\dfrac{1}{4}-\dfrac{3}{2}\end{matrix}\right.=\left[{}\begin{matrix}-x-\dfrac{5}{8}\\x-\dfrac{15}{8}\end{matrix}\right.\)
\(2.\left|\dfrac{5}{8}-x\right|=\dfrac{5}{4}\\\left|\dfrac{5}{8}-x\right|=\dfrac{5}{8} \\ \left[{}\begin{matrix}\dfrac{5}{8}-x=\dfrac{5}{8}\\\dfrac{5}{8}-x=-\dfrac{5}{8}\end{matrix}\right.\left[{}\begin{matrix}x=0\\x=\dfrac{5}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\left[{}\begin{matrix}x=4\\x\neg-\dfrac{3}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
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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:))
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a: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{BAD}\) chung
Do đó: ΔABD=ΔACE
Suy ra: BD=CE
b: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có
BC chung
EC=DB
Do đó:ΔEBC=ΔDCB
Suy ra: \(\widehat{KCB}=\widehat{KBC}\)
hay ΔKBC cân tại K
d: Xét ΔABK và ΔACK có
AB=AC
BK=CK
AK chung
Do đó: ΔABK=ΔACK
Suy ra: \(\widehat{BAK}=\widehat{CAK}\)
hay AK là tia phân giác của góc BAC
![](https://rs.olm.vn/images/avt/0.png?1311)
Hướng dẫn thoai , bìa này tư duy đc :))
a) các cặp song song là : QE//PH;PQ//FH
b)
Góc EOQ = 90o - góc OQE = 90o- 55o = 35o
Góc OFH = 90o - góc EOQ = 90o - 35o = 55o
Góc PHF = 90o - góc OFH = 90o - 55o = 35o
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a: \(=2x^2-3x+1+3x^2+2x-1=5x^2-x\)
b: \(=4x^3-2x^2+3x-2x^3-3x^2+4x=2x^3-5x^2+7x\)
c: \(=x^2-5x+6-3x^2+2x-1=-2x^2-3x+5\)
d: \(=2x^3+5x^2-3x+1-x^3+2x^2-x+1\)
\(=x^3+7x^2-4x+2\)
e: \(=3x^2+2x-4+4x^2-x+5=7x^2+x+1\)
f: \(=x^3-2x^2+5x-1-2x^3-3x^2+4x-2=-x^3-5x^2+9x-3\)
g: \(=4x^4-3x^3+x^2+2x-1+2x^3-4x^2+3x-1\)
\(=4x^4-x^3-3x^2+5x-2\)
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