Tìm y biết:
a/y:6=2010:15 b/ x+1/2=3/4 c/x-1/3=1/4 d/3x+3/8=1/2
e/ 5.y - 1952=2500-1947 f/(8.y - 1942). 1947=(240 - 194,2). 19470
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Ta có: (8y-1942).1947=(2400-1942).19470
\(\Rightarrow\)8y-1942=458.10
\(\Rightarrow\)8y=4580+1942
\(\Rightarrow\)8y=6522
\(\Rightarrow\)y=6522:8=815,25
\(\left(8.y-1942\right).1947=\left(2400-1942\right).19470\)
\(\Leftrightarrow\left(8.y-1942\right).1947=458.19470\)
\(\Leftrightarrow\left(8.y-1942\right).1947=8917260\)
\(\Leftrightarrow8.y-1942=8917260:1947\)
\(\Leftrightarrow8.y-1942=4580\)
\(\Leftrightarrow8.y=4580+1942\)
\(\Leftrightarrow8.y=6522\)
\(\Leftrightarrow y=6522:8\)
\(\Leftrightarrow y=\frac{3261}{4}\)
~ Rất vui vì giúp đc bn ~
=a, \(\dfrac{x}{15}\) = \(\dfrac{2}{5}\)
= \(x.5=15.2\)
=> \(x=\dfrac{15.2}{5}\)\(=\dfrac{30}{5}\) \(=6\)
Vậy \(x=6\)
b, \(\dfrac{3}{x-7}\) \(=\dfrac{27}{135}\)
= \(\dfrac{3}{x-7}\) \(=\dfrac{3}{15}\)
= \(x-7=15\)
\(x=15+7\)
\(x=22\)
vậy x = 22
c, \(320.x-10=5.48:24\)
= \(320x-10=240:24\)
= \(320x-10=10\)
= \(320x=10+10\)
\(320x=20\)
\(x=20:320\)
\(x=0,0625\)
d, \(5x-1952=\) \(2500-1947\)
\(5x-1952=553\)
\(5x=553+1952\)
\(5x=2505\)
\(x=2505:5\)
\(x=501\)
e, \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)\left(x+5\right)=45\)
= \(\left(x+x+x+x+x\right)\)+\(\left(1+2+3+4+5\right)\) \(=45\)
= \(5x+15=45\)
\(5x=45-15\)
\(5x=30\)
\(x=30:5\)
\(x=6\)
f, \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}-\dfrac{2}{63}=\dfrac{1}{9}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}=\dfrac{1}{9}+\dfrac{2}{63}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}-\dfrac{2}{35}=\dfrac{1}{7}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}=\dfrac{1}{7}+\dfrac{2}{35}\)
= \(x-\dfrac{2}{3}-\dfrac{2}{15}=\dfrac{1}{5}\)
= \(x-\dfrac{2}{3}=\dfrac{1}{5}+\dfrac{2}{15}\)
= \(x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\) \(\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=1\)
k, \(\dfrac{3+5+7+...+2015}{2+4+6+...+2014+x}=1\)
ta thấy phần tử là tập hợp các số lẻ ; phần mẫu là tập hợp các số chẵn
mà số chẵn hơn số lẻ 1 đơn vị
nên x thuộc tổng các số phần tử hơn mẫu là 1 đơn vị
=> từ \(2+4+6+...+2014\)có số số hạng là :
( 2014 - 2 ) : 2 + 1 = 1007
vậy x sẽ bằng :
( 1 + 1 ) . 1007 : 2 = 1007
vập số cần tìm là : 1007
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
x+1 | 1 | -1 | 5 | -5 |
y-2 | 5 | -5 | 1 | -1 |
x | 0 | -2 | 4 | -6 |
y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
x-5 | 1 | -1 | 7 | -7 |
y+4 | -7 | 7 | -1 | 1 |
x | 6 | 4 | 12 | -2 |
y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
a) \(\frac{y}{6}=\frac{2010}{15}\) c) \(x-\frac{1}{3}=\frac{1}{4}\) e)\(5y-1952=2500-1947\)
\(y=\frac{2010}{15}.6\) \(x=\frac{1}{4}+\frac{1}{3}\) \(5y-1952=553\)
\(y=804\) \(x=\frac{7}{12}\) \(5y=553+1952\)
\(5y=2505\)
\(y=2505:5=501\)
b) \(x+\frac{1}{2}=\frac{3}{4}\) c) \(3x+\frac{3}{8}=\frac{1}{2}\)
\(x=\frac{3}{4}-\frac{1}{2}\) \(3x=\frac{1}{2}-\frac{3}{8}\)
\(x=\frac{1}{4}\) \(3x=\frac{1}{8}\)
\(x=\frac{1}{8}:3\)
\(x=\frac{1}{24}\)
f)\(\left(8y-1942\right).1947=\left(240-194,2\right).19470\)
\(\left(8y-1942\right).1947=45,8.19470\)
\(\left(8y-1942\right)=45,8.19470:1947\)
\(8y-1942=45,8.10\)
\(8y-1942=458\)
\(8y=458+1942\)
\(8y=2400\)
\(y=2400:8\)
\(y=300\)