x+1 x+8
-----+------- giải hộ mình với nha :>>>>>>>>
6 3
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
4x5x6/3x10x8=120/240=120:120/240:120=1/2 nha
~HT~
k cho mình nha
@@@@@@@@@@@@@@@@
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\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{11\times12}\)
\(=\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+...+\frac{12-11}{11\times12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}\)
\(=\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: |x+3| \(\ge\)0; |2x+y-4| \(\ge\)0
\(\Rightarrow\) |x + 3| + |2x + y - 4| \(\ge\) 0
Dấu = xảy ra khi x+3=0 và 2x+y-4 = 0 \(\Rightarrow\)x=-3; y=10
1) |x + 3| + |2x + y - 4| = 0
\(\Leftrightarrow\hept{\begin{cases}x+3=0\\2x+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-6+y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=10\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> x-3 va 8-x trai dau
sau do xet th1: x-3>0 va 8-x<0
th2: x-3<0 va 8-x >0
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{1}{2}-\frac{1}{3}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\)
\(\frac{1}{4}\cdot\frac{1}{5}=\frac{1}{20}\)
\(\frac{1}{9}+\frac{1}{8}=\frac{8}{72}+\frac{9}{72}=\frac{17}{72}\)
\(\frac{1}{2}:\frac{1}{8}=\frac{1}{2}\cdot8=4\)
1/2 - 1/3 = 1/6
1/4 x 1/5 = 1/20
1/9 + 1/8 = 17/72
1/2 : 1/8 = 8/2 = 4
Mn ơi ủng hộ mik nhé !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
ĐKXĐ : x ≠ 1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 6
pt <=> \(\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{3x^2-9x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)
<=> \(\frac{6x^2-22x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)
=> \(\left(x-6\right)\left(6x^2-22x+18\right)=6\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
(bạn tự khai triển rút gọn nhé)
<=> \(6x^3-58x^2+150x-108=6x^3-36x^2+66x-36\)
<=>\(6x^3-58x^2+150x-108-6x^3+36x^2-66x+36=0\)
<=> \(-22x^2+84x-72=0\)
<=> \(11x^2-42x+36=0\)
(pt này lên lớp 9 mới học nên mình dừng tại đây)
\(\dfrac{x+1}{6}+\dfrac{x+8}{3}=\dfrac{x+1}{6}+\dfrac{2.\left(x+8\right)}{6}=\dfrac{3x+9}{6}\\ =\dfrac{3.\left(x+3\right)}{6}=\dfrac{x+3}{2}\)