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a: =>y/15=-2/3
hay y=-10
b: 2/x=x/18
nên \(x^2=36\)
hay \(x\in\left\{6;-6\right\}\)
c: x/9=16/x
nên \(x^2=144\)
hay \(x\in\left\{12;-12\right\}\)
`A)2/3=x/60`
`=>40/60=x/60`
`=>x=40`
`B)-1/2=y/18`
`=>-9/18=y/18`
`=>y=-9`
`C)3/x=y/35=-36/84`
Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`
`=>3/x=-3/7`
`=>x=-7`
`y/35=-3/7=-15/35`
`=>y=-15`
`D)7/x=y/27=-42/54`
Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`
`=>7/x=-7/9`
`=>x=-9`
`y/27=-7/9=-21/27`
`=>y=-21`
Giải:
a) \(y^2=3-\left|2x-3\right|\)
Vì \(-\left|2x-3\right|\le0\forall x\) nên \(3-\left|2x-3\right|\le3\forall x\) nên \(y^2\le3\rightarrow y^2\in\left\{0;1\right\}\) (vì \(y\in Z\) )
TH1:
\(y^2=0\)
\(\Rightarrow y=0\)
\(\Rightarrow\left|2x-3\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
TH2:
\(y^2=1\)
\(\Rightarrow y=\pm1\)
Lời giải:
a.
$(-2)x-(-21)=15$
$-2x+21=15$
$-2x=15-21=-6$
$x=(-6):(-2)=3$
b.
$(3x-2^2).7^3=7^4$
$3x-2^2=7^4:7^3=7$
$3x-4=7$
$3x=11$
$x=\frac{11}{3}$
a: x:(-9)=-54
=>\(x=\left(-54\right)\cdot\left(-9\right)\)
=>\(x=54\cdot9=486\)
b: \(x:\left(-12\right)=18\)
=>\(x=18\cdot\left(-12\right)=-216\)
c: \(x:\left(-5\right)=-19\)
=>\(x:5=19\)
=>\(x=19\cdot5=95\)
d: \(\left(x-28\right):\left(-12\right)=-5\)
=>\(x-28=\left(-12\right)\cdot\left(-5\right)=60\)
=>x=60+28=88
e: \(\left(x+15\right):\left(-28\right)=8\)
=>x+15=-28*8=-224
=>x=-224-15=-239
f: (x+30):(-45)=-4
=>\(x+30=\left(-45\right)\cdot\left(-4\right)=180\)
=>x=180-30
=>x=150
a) x : (-9) = -54
x= -54 . (-9)= 486
________
b) x : (-12) = 18
x= 18. (-12)= -216
_________
c) x : (-5) = -19
x= (-19). (-5)= 95
__________
d) (x - 28) : (-12) = -5
(x-28)= (-5). (-12)= 60
x= 60+28= 88
_______
e) (x + 15) : (-28) = 8
(x+15)= 8. (-28)= -224
x= -224 - 15 = - 239
__________
f) (x + 30) : (-45) = -4
(x+30)= -4. (-45)= 180
x= 180 - 30=150
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a: (x-3)(y+1)=15
=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)
=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}
=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}
b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=4+13\left(3^2+3^5+...+3^{98}\right)\)
=>m chia 13 dư 4
\(m=1+3+3^2+...+3^{99}+3^{100}\)
\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)
\(=1+40\left(3+3^5+...+3^{97}\right)\)
=>m chia 40 dư 1
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)
\(b.\)
\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)
\(c.\)
\(\dfrac{x}{4}=\dfrac{y}{5}\)
\(\Leftrightarrow\dfrac{2x}{8}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)
mà x+y=35
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(10;25)
b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
mà y-3x=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
Vậy: (x,y)=(1;5)
c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)
nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)
mà 2x-y=15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(20;25)
Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)