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b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
a.1+3+5+7+...+2x+1=225
[(2x+1-1):2+1]x(2x+1+1):2=225
(x+1)x(2x+2):2=225
(x+1)x(x+1)=225
(x+1)2=225
(x+1)2=152
x+1=15
x=14
_______________
b. 130-[5.(9-x)+43]=47
5.(9-x)+43=83
5.(9-x)=40
9-x=8
x=1
_______________
c.16x<324
24x<220
=>x∈{0;1;2;3;4}
a: =>(x+1)^2=225
=>x+1=15
=>x=14
b: =>[5*(9-x)+43]=130-47=83
=>5(9-x)=40
=>9-x=8
=>x=1
c: =>2^4x<2^20
=>4x<20
=>0<x<5
Bài 2:
Ta có: \(5^x.5^{x+2}\le10^{18}\div2^8\)
\(\Rightarrow5^{x+x+2}\le\left(10\div2\right)^{18}\)
\(\Rightarrow5^{2x+2}\le5^{18}\)
\(\Rightarrow2x+2\le18\Rightarrow2x\le16\Rightarrow x\le8\)
\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)
Bài 3:
Ta có: \(S=1+2+2^2+...+2^9=\left(2+2^2+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)
\(=2^{10}-1\left(1\right)\)
Ta có: \(5\times2^8=\left(2^2+1\right)\times2^8=2^{10}+2^8\left(2\right)\)
Từ (1) và (2) \(\Rightarrow S< 5\times2^8\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
Bài 1.
\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)
\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)
\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)
\(=1200:\left[450:\left(450-300\right)\right]\)
\(=1200:\left(450:150\right)\)
\(=1200:3\)
\(=400\)
\(---\)
\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)
\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)
\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)
\(=675-20\left[90-\left(164-100\right)\right]\)
\(=675-20\left(90-64\right)\)
\(=675-20\cdot26\)
\(=675-520\)
\(=155\)
\(---\)
\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)
\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)
\(=\left(1\cdot2022-1986\right)\cdot5-169\)
\(=\left(2022-1986\right)\cdot5-169\)
\(=36\cdot5-169\)
\(=180-169\)
\(=11\)
Bài 2.
\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)
Vậy \(x=2\).
\(---\)
\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)
Vậy \(x=\dfrac{52}{5}\).
\(Toru\)
Ý tưởng làm bài:
Nhận thấy: \(10^2=100\) (2 chữ số 0)
\(10^3=1000\) (3 chữ số 0)
\(10^4=10000\)(4 chữ số 0)
....v.v..
Suy ra \(10^n=100...00\) (có n chữ số 0)
Áp dụng vào bài:
b) sửa đề: \(2^x.2^{x+1}.2^{x+2}=100...000:5^{18}\) (có 18 chữ số 0)
Ta cần tìm x sao cho \(2^x.2^{x+1}.2^{x+2}=10^{18}:5^{18}\)
\(\Leftrightarrow2^{3x+3}=2^{18}\Leftrightarrow3x+3=18\Leftrightarrow3x=15\Leftrightarrow x=\frac{15}{3}=5\)
a)Sửa đề \(1+3+5+7+9+...+\left(2x+1\right)=242\) (thêm đk x > 0)
Dãy trên có số số hạng là: \(\frac{\left(2x+1-1\right)}{2}+1=x+1\) (số hạng)
Gọi tổng trên là A,ta có: \(A=\frac{\left(2x+1+1\right)\left(x+1\right)}{2}=\frac{2x+2}{2}.\frac{x+1}{2}\)
\(=\left(x+1\right).\frac{x+1}{2}=\frac{\left(x+1\right)^2}{2}=242\)
\(\Leftrightarrow\left(x+1\right)^2=484=22^2\Rightarrow x=21\)