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a: \(\left(\frac23-25\%\right)\cdot\left(-\frac25\right)^2-\left(1\frac25+1\frac13-2\right)\)
\(=\left(\frac23-\frac14\right)\cdot\frac{4}{25}-\left(1+\frac25+1+\frac13-2\right)\)
\(=\frac{5}{12}\cdot\frac{4}{25}-\left(\frac25+\frac13\right)\)
\(=\frac{20}{300}-\frac{11}{15}=\frac{1}{15}-\frac{11}{15}=-\frac{10}{15}=-\frac23\)
b: \(\left(\frac72+50\%\right)\cdot10-\left(3,6:2\frac25\right)^3\cdot\left(-\frac{2023}{2024}\right)^0\)
\(=\left(\frac72+\frac12\right)\cdot10-\left(\frac{18}{5}:\frac{12}{5}\right)^2\)
\(=4\cdot10-\left(\frac32\right)^2=40-2,25=37,75\)
a)(x − 12)2 = 0
=>x − 12 = 0
=> x = 12
b) (x+12)2 = 0,25
=> x + 12 = 0,5 hoặc x + 12= -0,5
=> x = -11,5 hoặc x = -12,5
c) (2x−3)3 = -8
=> 2x - 3 = -2
=> x = 0,5
d) (3x−2)5 = −243
=> 3x - 2 = -3
=> x = -1/3
e) (7x+2)-1 = 3-2
=> \(\dfrac{1}{7x+2}=\dfrac{1}{9}\)
=> 7x + 2 = 9
=> x = 1
f) (x−1)3 = −125
=> (x−1) = −5
=> x = -4
g) (2x−1)4 = 81
=> 2x - 1 = 3
=> x = 2
h) (2x−1)6 = (2x−1)8
=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1
=> x = 1/2 hoặc x = 1 hoặc x = 0
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ..
c/ \(\left(2x-3\right)^3=-8\)
\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-3=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
d/ \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
e/ \(\left(x-1\right)^3=-125\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x-1=-5\)
\(\Leftrightarrow x=-4\)
Vậy..
f/ \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy..
Lời giải:
a) \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)
\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow 2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
Cộng theo vế:
\(\Rightarrow B+2B=2^{201}-2\)
\(\Rightarrow B=\frac{2^{101}-2}{3}\)
c) Ta có:
\(C=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
\(\Rightarrow 3C=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
Cộng theo vế:
\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)
\(4C=3^{101}+1\Rightarrow C=\frac{3^{101}+1}{4}\)
a: \(3A=3+3^2+...+3^{101}\)
\(\Leftrightarrow2A=3^{101}-1\)
hay \(A=\dfrac{3^{101}-1}{2}\)
b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)
\(\Leftrightarrow3B=2^{101}-2\)
hay \(B=\dfrac{2^{101}-2}{3}\)
c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)
=>\(4C=3^{101}+1\)
hay \(C=\dfrac{3^{101}+1}{4}\)
\(3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot-\frac{85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\left(-\frac{17}{3}\right)-\frac{5}{6}\)
\(=\frac{-25}{6}\)
\(3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\frac{-85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3.\left(-85\right)}{5.9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{1.\left(-17\right)}{1.3}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{3}{1}-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{18}{6}-\frac{4}{6}+\frac{-34}{6}-\frac{5}{6}\)
\(=\frac{18-4+\left(-34\right)-5}{6}\)
\(=\frac{-25}{6}\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x+3y-z-5}{9}=\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) có 2x + 3y - z = 50
\(\Rightarrow\frac{50-5}{9}=5=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\hept{\begin{cases}x-1=10\\y-2=15\\z-3=20\end{cases}\Rightarrow\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}}\)
Trả lời:
Ta có:\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)\(=\frac{2x+3y-z-5}{9}\)(Tính chất dãy tỉ số bẳng nhau)
Mà\(2x+3y-z=50\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{50-5}{9}=\frac{45}{9}=5\)
\(\Rightarrow\hept{\begin{cases}2x-2=20\\3y-6=45\\z-3=20\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=22\\3y=51\\z=23\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\)
Vậy\(\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\)
Hok tốt!
Vuong Dong Yet
a: \(\left(\frac23-25\%\right)\cdot\left(-\frac25\right)^2-\left(1\frac25+1\frac13-2\right)\)
\(=\left(\frac23-\frac14\right)\cdot\frac{4}{25}-\left(1+\frac25+1+\frac13-2\right)\)
\(=\frac{5}{12}\cdot\frac{4}{25}-\left(\frac25+\frac13\right)\)
\(=\frac{20}{300}-\frac{11}{15}=\frac{1}{15}-\frac{11}{15}=-\frac{10}{15}=-\frac23\)
b: \(\left(\frac72+50\%\right)\cdot10-\left(3,6:2\frac25\right)^3\cdot\left(-\frac{2023}{2024}\right)^0\)
\(=\left(\frac72+\frac12\right)\cdot10-\left(\frac{18}{5}:\frac{12}{5}\right)^2\)
\(=4\cdot10-\left(\frac32\right)^2=40-2,25=37,75\)
= (2/3 - 1/4).4/25 - (7/5 + 4/3 - 2)
= ( 8/12 - 3/12).4/25-(21/15 + 20/15 - 30/15)
= 5/12.4/25 -11/15
=1/15 - 11/15
= -10/15
b, =(7/2 + 1/2).10 - (18/5.5/12) mũ 3 .(-1)
= 40 - 27/8 .(-1)
= 40 + 27/8 = 43,375