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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
a)|x+0,573|=2
=>x+0,573=2 hoặc -2
Xét x+0,573=2
=>x=1,427
Xét x+0,573=-2
=>x=-2,573
a) | x + 0,573 | = 2
\(\Rightarrow\)x + 0,573 = 2 hoặc x + 0,573 = -2
+) x + 0,573 = 2\(\Rightarrow\)x = 1,427
+) x + 0,573 = -2\(\Rightarrow\)x = -2,573
Vậy x = 1,427 hoặc -2,573
b) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow x+\frac{1}{3}=3\) hoặc \(x+\frac{1}{3}=-3\)
+) \(x+\frac{1}{3}=3\Rightarrow x=\frac{8}{3}\)
+) \(x+\frac{1}{3}=-3\Rightarrow x=\frac{-10}{3}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)
Các phần khác làm tương tự nhé bạn
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
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a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
Bài 2:
a) \(\frac{8^{14}}{4^{12}}\)
\(=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}\)
\(=\frac{2^{42}}{2^{24}}\)
\(=2^{18}\)
\(=262144.\)
b) \(\left(-\frac{1}{3}\right)^7.3^7\)
\(=\left[\left(-\frac{1}{3}\right).3\right]^7\)
\(=\left(-1\right)^7\)
\(=-1.\)
c) \(\frac{90^2}{15^2}\)
\(=\left(\frac{90}{15}\right)^2\)
\(=6^2\)
\(=36.\)
d) \(\frac{790^4}{79^4}\)
\(=\left(\frac{790}{79}\right)^4\)
\(=10^4\)
\(=10000.\)
Chúc bạn học tốt!
Mk làm tiếp cho bạn Vũ Minh Tuấn nhé!
Bài 1:
\(-\frac{64}{343}=x^3\)
\(\Rightarrow x^3=\left(-\frac{4}{7}\right)^3\)
\(\Rightarrow x=-\frac{4}{7}\)
Vậy \(x=-\frac{4}{7}\)
\(\left(x+20\right)^{100}+\left|y+4\right|=0\)
Ta có: \(\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy \(x=-20;y=-4\)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)