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2 tháng 4 2018
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}\)
\(=\left(\frac{3}{29}\cdot\frac{29}{3}\right)-\left(\frac{1}{5}\cdot\frac{29}{3}\right)\)
\(=1-\frac{29}{15}\)
\(=\frac{-14}{15}\)
b)\(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
=\(=\frac{16\cdot\left(-5\right)\cdot54\cdot56}{15\cdot14\cdot24\cdot21}\)
\(=\frac{2^4\cdot\left(-5\right)\cdot2\cdot3^3\cdot2^3\cdot7}{3\cdot5\cdot7\cdot2\cdot2^3\cdot3\cdot7}\)
\(=2^4\)
c)\(\frac{37}{7}\cdot\frac{8}{11}+\frac{37}{7}\cdot\frac{5}{11}-\frac{37}{7}\cdot\frac{2}{11}\)
\(=\frac{37}{7}\cdot\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
\(=\frac{37}{7}\cdot1\)
\(=\frac{37}{7}\)
Đúng nhớ k nhen!
21 tháng 5 2022
a: \(\dfrac{7}{11}< x-\dfrac{1}{7}< \dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{7}{11}+\dfrac{1}{7}< x< \dfrac{10}{13}+\dfrac{1}{7}\)
hay 60/77<x<83/91
b: \(\dfrac{7}{9}< \dfrac{13}{11}-x< \dfrac{15}{16}\)
\(\Leftrightarrow\dfrac{-7}{9}>x-\dfrac{13}{11}>-\dfrac{15}{16}\)
\(\Leftrightarrow-\dfrac{7}{9}+\dfrac{13}{11}>x>\dfrac{-15}{16}+\dfrac{13}{11}\)
\(\Leftrightarrow\dfrac{40}{99}>x>\dfrac{43}{176}\)
25 tháng 3 2017
11) \(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}=-\dfrac{7}{16}\)
\(\left(x-\dfrac{1}{2}\right)^2=-\dfrac{7}{16}+\dfrac{1}{2}\)
\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(x-\dfrac{1}{2}=\sqrt{\dfrac{1}{16}}\)
\(x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\)
25 tháng 3 2017
12) \(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3\right\}=5.21-89\\ \left\{x^2-\left[36-\left(64-63\right)^3-35\right]^3\right\}=105-89\\ \left\{x^2-\left[36-\left(1\right)^3-35\right]^3\right\}=16\\ \left\{x^2-\left[36-1-35\right]^3\right\}=16\\ \left\{x^2-0^3\right\}=16\\ x^2-0=16\\ x^2=16+0\\ x^2=16\\ x^2=4^2\\ \Rightarrow x=2\)
AS
2
13 tháng 1 2019
a) \(15+\left|x-2\right|=-41+11\)
\(15+\left|x-2\right|=-30\)
\(\left|x-2\right|=-45\)( vô lí )
b) \(\left|x+3\right|-8-16=-17\)
\(\left|x+3\right|=-17+8+16\)
\(\left|x+3\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}}\)
13 tháng 1 2019
a,\(15+\left|x-2\right|=-41+11\)
\(\Rightarrow\left|x-2\right|=-41+11-15\)
\(\Rightarrow\left|x-2\right|=-45\)
Vì \(\left|x\right|\ge0\Rightarrow\left|x-2\right|=-45\)(vô lí)
b,\(\left|x+3\right|-8-16=-17\)
\(\Rightarrow\left|x+3\right|=-17+16+8\)
\(\Rightarrow\left|x+3\right|=7\)
=> x+3= 7 hoặc x+3=-7
=> x=4 hoặc x=-10
XO
12 tháng 8 2020
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}..1\frac{1}{99}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{100}{99}=\frac{2.2.3.3.4.4...10.10}{1.3.2.4.3.5...9.11}=\frac{\left(2.3.4...10\right)\left(2.3.4...10\right)}{\left(1.2.3...9\right)\left(3.4.5...11\right)}\)
\(\frac{10.2}{1.11}=\frac{20}{11}\)
b) \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}\)
\(=\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(1.2.3.4.5\right).\left(3.4.5.6.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{1.7}{6.2}=\frac{7}{12}\)
c) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(-\frac{98}{97}+\frac{1}{97}\right)=1-1=0\)
d) \(3\frac{1}{11}.\frac{27}{36}.1\frac{6}{7}.2\frac{4}{9}=\frac{34}{11}.\frac{3}{4}.\frac{13}{7}.\frac{22}{9}=\frac{34.3.13.22}{11.4.7.9}=\frac{34.13}{11.2.7.3}=\frac{442}{462}=\frac{221}{231}\)
16 tháng 8 2018
34 +14 :x=−2
\(\frac{1}{4}:x=-2-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-11}{4}\)
\(x=\frac{1}{4}:\frac{-11}{4}\)
\(x=\frac{-1}{11}\)
x2:1611 =114
tìm x
KD
16 tháng 8 2018
\(\frac{3}{4}+\frac{1}{4}:x=-2\)
\(\frac{1}{4}:x=-2-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-11}{4}\)
\(x=\frac{1}{4}:\frac{-11}{4}\)
\(x=\frac{-1}{11}\)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(x^2=\frac{11}{4}\times\frac{16}{11}\)
\(x^2=4\)
\(\Rightarrow x=-2\)\(\text{hoặc}\)\(x=2\)
vì x + 16 ⋮ x + 1
x+1⋮ x+1
=>(x+16)-(x+1)\(⋮x+1\)
=>\(15⋮x+1\)
=>(x+1)\(\inƯ\left(15\right)\) ={\(\pm1;\pm3;\pm5;\pm15\) }
ta có bảng
vậy x\(\in\left\{-16;-6;-4;-2;0;2;4;14\right\}\)
x + 11 ⋮ x + 1
vì \(x+1⋮x+1\)
=>\(\left(x+11\right)-\left(x+1\right)⋮\left(x+1\right)\)
=>\(\left(x+11-0x-1\right)⋮\left(x+1\right)\)
=> \(10⋮x+1\)
=>\(x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
ta có bảng sau:
9
vậy\(x\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)