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\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
1. thực hiện phép tính
a, 23. 15 - [ 115 - ( 12-5)2 ]
= 23 . 15 - [ 115 - 72 ]
= 8 . 15 - 66
= 120 - 66
= 54
b,132 - [ 116 - (132 - 128)2
= 132 - [ 116 - 42 )
= 132 - 100
= 32
c, [ 545 - ( 45 + 4.25 ) ] : 50 - 2000: 250 +215: 213
= [ 545 - 145 ] : 50 -8 + 22
= 400 : 50 - 8 + 4
= 8 - 8 + 4
= 4
d, [ 1104 - ( 25.8 + 40)] :9 + 316: 312
= [ 1104 - { 200+40 } ] : 9 + 34
= { 1104 - 240 ) : 9 + 81
= 864 : 9 + 81
= 177
2.tìm x bt
a, 575 - ( 6x + 70) = 445
=> 6x +70 = 575 - 445
=> 6x + 70 = 130
=> 6x = 130 - 70
=> 6x = 60
=> x = 60:6
=> x = 10
Vậy x = 10
b, 315 + (125 - x) = 435
=> 125 - x = 435-315
=> 125-x = 120
=> x = 125-120
=> x = 5
Vậy x = 5
c, (3-x).(x-3)=0
=> \(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=3-0\\x=0+3\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=3\end{cases}}\)
Vậy x = 3
a,
3x + 3 - [7x+4] = 7 + [4x-1]
=> 3x + 3 - x - 4 = 7 + 4x - 1
=> 2x - 1 = 6 + 4x
=> 2x - 4x = 6 + 1
=> -2x = 7
=> x = -7/2
b,
3x+1 + 3x+3 =810
=> 3x+1[1 + 32] = 810
=> 3x+1 = 810 / 10
=> 3x+1 = 81
=> x = 4
c, \(1\frac{1}{2}:\left[\frac{1}{2}-\frac{1}{3}\right]-x=5\)
\(\Rightarrow\frac{3}{2}:\frac{1}{6}-x=5\Leftrightarrow9-x=5\)
\(\Leftrightarrow x=4\)
d,
\(2,4:\left[25\%+\frac{x}{40}\right]-\frac{12}{15}=3\frac{1}{5}\)
\(\Rightarrow\frac{12}{5}:\left[\frac{1}{4}+\frac{x}{40}\right]-\frac{12}{15}=\frac{16}{5}\)
\(\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=\frac{16}{5}+\frac{12}{15}\Leftrightarrow\frac{12}{5}:\left[\frac{10}{40}+\frac{x}{40}\right]=4\)
\(\Rightarrow\frac{10+x}{40}=\frac{12}{5}:4\Leftrightarrow\frac{10+x}{40}=\frac{3}{5}\)
\(\Rightarrow\frac{10+x}{40}=\frac{24}{40}\Leftrightarrow10+x=24\Rightarrow x=14\)
a) 3x + 3 - ( x + 4 ) = 7 + ( 4x - 1 )
3x + 3 - x - 4 = 7 + 4x - 1
2x - 1 = 6 + 4x
-2x = 7
\(\Rightarrow\)x = \(\frac{-7}{2}\)
b) 3x+1 + 3x+3 = 810
3x . 3 + 3x . 33 = 810
3x . ( 3 + 33 ) = 810
3x . 30 = 810
3x = 810 : 30
3x = 27
3x = 33
\(\Rightarrow\)x = 3
c) \(1\frac{1}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\left(\frac{1}{2}-\frac{1}{3}\right)-x=5\)
\(\frac{3}{2}:\frac{1}{6}-x=5\)
\(9-x=5\)
\(\Rightarrow x=9-5\)
\(\Rightarrow x=4\)
d) 2,4 : ( 25% + \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(3\frac{1}{5}\)
\(\frac{12}{5}\) : ( \(\frac{1}{4}\)+ \(\frac{x}{40}\)) - \(\frac{12}{15}\)= \(\frac{16}{5}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=\frac{16}{5}+\frac{12}{15}\)
\(\frac{12}{5}:\left(\frac{1}{4}+\frac{x}{40}\right)=4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{12}{5}:4\)
\(\frac{1}{4}+\frac{x}{40}=\frac{3}{5}\)
\(\frac{x}{40}=\frac{3}{5}-\frac{1}{4}\)
\(\frac{x}{40}=\frac{7}{20}\)
\(\Rightarrow\frac{x}{40}=\frac{14}{40}\)
\(\Rightarrow x=14\)
a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)
b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)
c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)
d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
\(a,x^2=4\Rightarrow x^2=2^2\Rightarrow x=2\)
\(b,x^2=64\Rightarrow x^2=8^2\Rightarrow x=8\)
\(c,6x^3-8=40\Rightarrow6x^3=48\Rightarrow x^3=8\Rightarrow x^3=2^3\Rightarrow x=2\)
\(d,\left(2x-1\right)^2=49\Rightarrow\left(2x-1\right)^2=7^2\Rightarrow2x-1=7\Rightarrow x=4\)
\(e,2^x:16=2^5\Rightarrow2^x:16=32\Rightarrow2^x=512\Rightarrow2^x=2^9\Rightarrow x=9\)
\(f,4^5:4^x=16\Rightarrow1024:4^x=16\Rightarrow4^x=64\Rightarrow4^x=4^3\Rightarrow x=3\)
a, x^2 = 4
=> x = 2 hoặc x = -2
b, x^2 = 64
=> x = 8 hoặc x = -8
c, 6x^3 - 8 = 40
=> 6x^3 = 48
=> x^3 = 8
=> x = 2
d, (2x - 1)^2 = 49
=> 2x - 1 = 7 hoặc 2x - 1 = -7
=> 2x = 8 hoặc 2x = -6
=> x = 4 hoặc x = -3
e, 2^x : 16 = 2^5
=> 2^x : 2^4 = 2^5
=> 2^x = 2^9
=> x = 9
f, 4^5 : 4^x = 16
=> 4^5 - x = 4^2
=> 5 - x = 2
=> x = 3
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
a) Ta có: \(x^4\ge0\) \(\forall x\)
\(\left(y-2\right)^2\ge0\) \(\forall y\)
\(\Rightarrow A\ge-8\). Dấu = khi <=> \(\hept{\begin{cases}x^4=0\\\left(y-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy min A = -8 <=> \(\hept{\begin{cases}x=0\\y=2\end{cases}}\)
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