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a) \(x-270:45=120\)
\(x-270=120.45\)
\(x-270=5400\)
\(x=5400+270\)
\(x=5670\)
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b) \(575-\left(6\times x+70\right)=445\)
\(6\times x+70=575-445\)
\(6\times x+70=130\)
\(6\times x=130-70\)
\(6\times x=60\)
\(x=60:6\)
\(x=10\)
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c) \(x-105:21=15\)
\(x-105=15.21\)
\(x-105=315\)
\(x=315+105\)
\(x=420\)
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d) \(3200:\left(405-x\right)=8\)
\(405-x=3200:8\)
\(405-x=400\)
\(x=405-400\)
\(x=5\)
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i) \(43575-75\times x=42450\)
\(75.x=43575-42450\)
\(75.x=1125\)
\(x=1125:75\)
\(x=15\)
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e) \(8,75\times x+1,25\times x=20\)
\(x\times\left(8,75+1,25\right)=20\)
\(x\times10=20\)
\(x=20:10\)
\(x=2\)
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\(A=\dfrac{\left|x-150\right|+102}{\left|x-150\right|+100}=\dfrac{\left|x-150\right|+100+2}{\left|x-150\right|+100}=1+\dfrac{2}{\left|x-150\right|+100}\)
Có: |x - 150| ≥ 0
=> |x - 150| + 100 ≥ 100
\(\Rightarrow\dfrac{2}{\left|x-150\right|+100}\le\dfrac{1}{50}\)
\(\Rightarrow A\le\dfrac{51}{50}\)
Dấu = xảy ra khi x = 150
Vậy:...
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\(A=\dfrac{\left|x-150\right|+102}{\left|x-150\right|+100}\\ A=\dfrac{\left|x-150\right|+100+2}{\left|x-150\right|+100}\\ A=1+\dfrac{2}{\left|x-150\right|+100}\)
\(\left|x-150\right|\ge0\forall x\\ \Rightarrow\left|x-150\right|+100\ge100\forall x\\ \Rightarrow\dfrac{2}{\left|x-150\right|+100}\le\dfrac{1}{50}\forall x\\ \Rightarrow A=1+\dfrac{2}{\left|x-150\right|+100}\le\dfrac{51}{50}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x-150\right|=0\\ \Leftrightarrow x-150=0\\ \Leftrightarrow x=150\)
Vậy GTLN của \(A=\dfrac{51}{50}\) khi x = 150
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\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)
Vậy \(X=\frac{147}{50}\)
( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......+ 1/9 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89 . 1/2
( 1 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89/2
90 - 5/2 : ( x + 103/50 ) = 89/2
5/2 : ( x + 103/50 ) = 90 - 89/2
5/2 : ( x + 103/50 ) = 91/2
x + 103/50 = 5/2 : 91/2
x + 103/50 = 5/91
x = 5/91 - 103/50
x = -9,123/4550
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x.3,7+x.6,3=120b,x.3,7+x.6,3=120
x(3,7+6,3)=120x(3,7+6,3)=120
10x=12010x=120
x=120:10x=120:10
x=12(tm)
(15×24-x):0.25=100 : 1/4
(15×24-x):0.25 = 400
(15x24-x) = 100
360-x =100
x=260
(8.75+1.25 +5+1) x X=20
16 x X=20
X=20 :16
X=1.25
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Quy đồng, ta được:
\(\dfrac{0,8:\left(\dfrac{4}{5}.\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{500-2}{5}\right):\dfrac{4}{7}}{6\left(\dfrac{5}{9}-\dfrac{13}{4}\right).\dfrac{36}{17}}\)
\(=\dfrac{0,8:1}{\dfrac{15}{25}}+\dfrac{\dfrac{498}{5}.\dfrac{7}{4}}{6\left(\dfrac{20-117}{36}\right).\dfrac{36}{17}}\)
\(=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}+\dfrac{\dfrac{1743}{10}}{6.\dfrac{-97}{36}.\dfrac{36}{17}}\)
\(=\dfrac{4}{3}+\dfrac{\dfrac{1743}{10}}{\dfrac{-582}{17}}\)
\(=\dfrac{4}{3}-\dfrac{9877}{1940}=\dfrac{4.1940-9877.3}{3.1940}=\dfrac{-21871}{5820}\)
Số to quá !!!!
=> |x-1,25| = 1
=> x -1,25 = 1 hoặc -1
=> x = 2,25 hoặc 0,25
x - 1,25 = 100%
x - 1,25 = 1
x = 1 + 1,25
x = 2,25