\(25\%\cdot x+x=-1,25\)
\(75\%\cdot x+x:\dfrac{4}{5}=-2010\)
\(25\%\cdot x+0,5\cdot x-\dfrac{x}{3}=-26\)
giúp mink vs nha thanks 
Những câu hỏi liên quan
1: rút gọn rồi tính
left(-dfrac{72}{40}-dfrac{144}{60}-2dfrac{1}{3}right) : left(dfrac{45}{100}-dfrac{25}{60}+-dfrac{75}{25}right)
2: tìm x: 3cdotleft(4-xright)+left(x+2right)cdotleft(1+2xright)7cdotleft(1+xright)-2xcdotleft(2-xright)
3: tìm x: dfrac{2cdotleft(1+xright)}{3}-dfrac{5cdotleft(2-xright)}{6}1dfrac{1}{3}-dfrac{3cdotleft(2x+3right)}{4}-1dfrac{1}{2}cdotleft(x+1right)
4: cho a 3+3^{2^3}+3^3+3^4+...+3^{360}
Đọc tiếp
1: rút gọn rồi tính
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right)\) : \(\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
2: tìm x: \(3\cdot\left(4-x\right)+\left(x+2\right)\cdot\left(1+2x\right)=7\cdot\left(1+x\right)-2x\cdot\left(2-x\right)\)
3: tìm x: \(\dfrac{2\cdot\left(1+x\right)}{3}-\dfrac{5\cdot\left(2-x\right)}{6}=1\dfrac{1}{3}-\dfrac{3\cdot\left(2x+3\right)}{4}-1\dfrac{1}{2}\cdot\left(x+1\right)\)
4: cho a= \(3+3^{2^3}+3^3+3^4+...+3^{360}\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
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Tìm x, biết:
\(x=\dfrac{19}{11}\cdot\dfrac{5}{14}+\dfrac{1}{11}\cdot\dfrac{5}{7}-\sqrt{\dfrac{25}{4}}\cdot\dfrac{2}{11}\)
tìm x:
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
Ta có:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3.....9}{2.3.4.....10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
\(\Leftrightarrow x=201\)
Vậy x = 201
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\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{10}\right)=\)\(\dfrac{x}{2010}\)
\(\Leftrightarrow\) \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}......\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3......9}{2.3.4.......10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow2010=10x\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
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1,Tìm x, biết:
a/\(^{\left(x-1\right)^{x+2}}=^{\left(x-1\right)^{x+4}}\)
b/\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
c/\(4^x+3^x=2^x+6^x\left(STN\right)\)
GIÚP MÌNH NHANH NHA MÌNH TICK CHO
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
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Tìm x, biết
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)=\(4^x\)
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)
Xét mẫu số:
\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)
\(=2^{32}(1.2.3....31.32)\)
Suy ra:
\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)
Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)
\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)
Vậy \(x=\frac{-37}{2}\)
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Tìm x biết :
\(\dfrac{3}{\left(x+2\right)\cdot\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\cdot\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\cdot\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\cdot\left(x+17\right)}\) Vs x \(\notin\){-2;-5;-17;-10}
\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
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1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
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Giải các bất phương trình :
a) \(\dfrac{x^3-4\cdot x^2+5\cdot x-20}{x^3-x^2-10\cdot x-8}>0\)
b) \(\dfrac{x^2+2\cdot x+2}{x+1}>\dfrac{x^2+4\cdot x+5}{x+2}-1\)
Làm từng câu cũng được nha các bạn!
Xin cám ơn!
b) \(\dfrac{x^2+2\cdot x+2}{x+1}>\dfrac{x^2+4\cdot x+5}{x+2}-1\)
\(\Leftrightarrow\dfrac{x^2+2\cdot x+2}{x+1}-\dfrac{x^2+4\cdot x+5}{x+2}+1>0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x+1\right)\left(x^2+4x+5\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+4x^2+5x+x^2+4x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-\left(x^3+5x^2+9x+5\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+2x^2+2x+2x^2+4x+4-x^3-5x^2-9x-5+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{0+0+1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)>0\)
\(\left\{{}\begin{matrix}x+1>0\\x+2>0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+1< 0\\x+2< 0\end{matrix}\right.\)
↓
\(\left\{{}\begin{matrix}x>-1\\x>-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< -1\\x< -2\end{matrix}\right.\)
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\(\dfrac{4x-8}{x+5}\cdot\dfrac{25-x^2}{2x-x^2}\)
\(=\dfrac{4\left(x-2\right)}{x+5}\cdot\dfrac{\left(x+5\right)\left(x-5\right)}{x\left(x-2\right)}=\dfrac{4x-20}{x}\)
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\(\dfrac{4x-8}{x+5}.\dfrac{25-x^2}{2x-x^2}=\dfrac{4\left(x-2\right)}{x+5}.\dfrac{\left(x+5\right)\left(x-5\right)}{x\left(x-2\right)}=\dfrac{4\left(x-5\right)}{x}\)
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