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Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{32}{91}\)
\(x=\frac{32}{91}+2\)
\(x=\frac{212}{91}\)
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
$\textbf{b)}$
$\text{Điều kiện: }x\ne1,\ 3,\ 8,\ 20.$
$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$
$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$
$\Leftrightarrow\dfrac1{x-1}=\dfrac34$
$\Leftrightarrow4=3(x-1)$
$\Leftrightarrow3x=7$
$\Leftrightarrow x=\dfrac73.$
$\textbf{a)}$
Điều kiện: $x\ne-2,\,-5,\,-10,\,-17.$
$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\left(\dfrac1{x+2}-\dfrac1{x+5}\right)+\left(\dfrac1{x+5}-\dfrac1{x+10}\right)+\left(\dfrac1{x+10}-\dfrac1{x+17}\right)=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac{x+17-x-2}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$
$\Leftrightarrow x=15.$
$\textbf{a)}$
$\dfrac{x-1}{2}+\dfrac{x-2}{5}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{5(x-1)+2(x-2)}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow\dfrac{7x-9}{10}=\dfrac14+\dfrac{x-7}{10}$
$\Leftrightarrow14x-18=5+2x-14$
$\Leftrightarrow12x=9$
$\Leftrightarrow x=\dfrac34.$
$\textbf{b)}$
$\dfrac{3-2}{2x-3}=\dfrac25+\dfrac1{2x-3}-\dfrac32$
$\Leftrightarrow1-\dfrac2{2x-3}=-\dfrac{11}{10}+\dfrac1{2x-3}$
$\Leftrightarrow\dfrac{21}{10}=\dfrac3{2x-3}$
$\Leftrightarrow21(2x-3)=30$
$\Leftrightarrow42x=93$
$\Leftrightarrow x=\dfrac{31}{14}.$
a) 3. ( x - 2/5 ) = 0
<=> x = 0 : 3 + 2/5
<=> x = 2/5
( Theo cách lớp 6)
Hoặc cách thông thường nek:
3. ( x - 2/5 ) = 0
x - 2/5 = 0 : 3
x - 2/5 = 0
x = 0 + 2/5
x = 2/5
Mấy câu sau làm như vậy nha!
^^ Chúc học tốt!!!!
\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)
\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)
ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP