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\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)\left(1-\dfrac{1}{21}\right)\left(1-\dfrac{1}{28}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{6}{6}-\dfrac{1}{6}\right)\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\left(\dfrac{15}{15}-\dfrac{1}{15}\right)\left(\dfrac{21}{21}-\dfrac{1}{21}\right)\left(\dfrac{28}{28}-\dfrac{1}{28}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.\dfrac{14}{15}.\dfrac{20}{21}.\dfrac{27}{28}\)
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
Câu 1:
c: 2x=3y
nên x/3=y/2
=>x/9=y/6
5y=3z
nên y/3=z/5
=>y/6=z/10
=>x/9=y/6=z/10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{3x+3y-7z}{3\cdot9+3\cdot6-7\cdot10}=\dfrac{35}{-25}=-\dfrac{7}{5}\)
Do đó: x=-63/5; y=-42/5; z=-14
Bài 2:
Gọi ba số lần lượt là a,b,c
Theo đề, ta có: 4/3a=b=3/4c
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{4}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{4}{3}}\)
\(\Leftrightarrow\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}\)
Đặt \(\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}=k\)
=>a=9k; b=12k; c=16k
Theo đề, ta có: \(a^2+b^2+c^2=481\)
\(\Leftrightarrow81k^2+144k^2+256k^2=481\)
=>k2=1
Trường hợp 1: k=1
=>a=9; b=12; c=16
Trường hợp 2: k=-1
=>a=-9; b=-12; c=-16
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ 
Giải:
a) \(\dfrac{1}{4}+x-\dfrac{1}{4}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}+\dfrac{3}{4}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{4}x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy ...
b) \(\left|x^2-2x\right|+\left|x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x^2-2x\right|=0\\\left|x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow x=0\)
Vậy ...
c) \(\left|3x^2-2x\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=x\\3x^2-2x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2=3x\\3x^2=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-3x=0\\3x^2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x\left(x-1\right)=0\\x\left(3x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
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 \(x=\frac{-37}{2}\)
a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)
b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10
a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)
c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)
d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)
e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)
= 390326,9437