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Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
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) =
1
- fddfssdfdsfdssssssssssssssffffffffffffffffffsssssssssssssssssssfsssssssssssssssssssssssfffffffffffffff
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: \(xy\ge0\)
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
4^5=2^10
9^4=3^8
2*6^9=2^10*3^9
thì cái tử sẽ đc:
2^10*(-3)
mẫu e phân tích tt