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a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow82< x< 92\)
\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)
b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)
\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)
\(\Rightarrow-1\le x\le7\)
\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)
Ta có :\(\frac{x-1}{-2}=\frac{-8}{x-1}\Rightarrow\left(x-1\right)^2=16\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
\(\frac{x-1}{-2}=\frac{-8}{x-1}\left(x\ne1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-2\cdot\left(-8\right)\)
\(\Leftrightarrow\left(x-1\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}\left(tm\right)}}\)
Vậy x=5; x=-3
Bài 3
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=25\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow\left(-x\right).x=\left(-9\right).4\)
\(\Rightarrow-x=-36\)
\(\Rightarrow x=36\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x.\left(x+1\right)=4.18\)
\(\Rightarrow x.\left(x+1\right)=72\)
Vì x và x + 1 là 2 số tự nhiên liên tiếp
\(\Rightarrow x\left(x+1\right)=8.9\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=8\end{cases}}\)
Bài 4
\(\frac{x-4}{y-3}=\frac{4}{3},x-y=5\)
Ta có :
\(x-y=5\)
\(\Rightarrow x=5+y\)
\(\Rightarrow\frac{y+5-4}{y-3}=\frac{4}{3}\)
\(\Rightarrow\frac{y+1}{y-3}=\frac{4}{3}\)\(\)
\(\Rightarrow\left(y+1\right).3=\left(y-3\right).4\)
\(\Rightarrow y.3+1.3=y.4-3.4\)
\(\Rightarrow y.3+3=y.4-12\)
\(\Rightarrow y.3-y.4=-12-3\)
\(\Rightarrow-1y=-15\)
\(\Rightarrow y=\left(-15\right):\left(-1\right)\)
\(\Rightarrow y=15\)
Vì x = y + 5
\(\Rightarrow x=15+4\)
\(\Rightarrow x=19\)
Vậy x = 19 , y = 15
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow\left(-x\right).x=4.\left(-9\right)\)
\(\Rightarrow-x=-9;x=4\)
\(\Rightarrow x=9;x=4\)
a)ta có xy=7*9=7*3*3
vậy x =9;21 , y=7;3
b) xy=-2*5
mà x<0<y
nên x=-2 ,y=5
c)x-y=5 hay x=y+5
\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)
a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
a/ \(\frac{x+2}{27}=\frac{x}{9}\)
=> 9(x + 2) = 27x
=> 9x + 18 = 27x
=> 9x + 18 - 27x = 0
=> 9x - 27x + 18 = 0
=> -18x = -18
=> x = 1
b/ \(\frac{-7}{x}=\frac{21}{34-x}\)
=> -7(34 - x) = 21x
=> -238 + 7x = 21x
=> 21x - 7x = -238
=> -14x = 238
=> x = -17
c) \(\frac{-8}{15}< \frac{x}{40}< \frac{-7}{15}\)
Ta có BCNN(15,40,15) = 120
=> \(\frac{-64}{120}< \frac{3x}{120}< \frac{-56}{120}\)
=> -64 < 3x < -56
=> x \(\in\){ -19;-20;-21}
Câu d tương tự
a; 3:\(\frac{2x}{5}\)= 1:0.001
3:\(\frac{2x}{5}\)=1000
\(\frac{2x}{5}\)=1000:3
\(\frac{2x}{5}\)=0.003
2x=0.003.5
2x=0.015
x=0.015:2
x=7.5
\(\frac{2}{x+1}=\frac{8}{x-2}\)
\(\Rightarrow2\left(x-2\right)=8\left(x+1\right)\)
\(\Rightarrow2x-4=8x+8\)
\(\Rightarrow2x-8x=8+4\)
\(\Rightarrow-6x=12\)
\(\Rightarrow x=12:\left(-6\right)\)
\(\Rightarrow x=-2\)
dễ