1/5x : 3= 2/9 : 0,25 tìm x,y
tìm x trong tỉ lệ thức 1/5x:3=2/3:0,25 là?
\(\frac{\frac{1}{5}x}{3}=\frac{\frac{2}{4}}{\frac{1}{4}}\)
\(\Leftrightarrow\frac{1}{5}x:3=2\)
\(\Rightarrow\frac{1}{5}x=6\)
\(\Rightarrow x=\frac{6}{\frac{1}{5}}\)
\(x=30\)
\(\frac{1}{5}x:3=\frac{2}{3}:0,25\)
\(\frac{1}{5}x:\frac{2}{3}=3:0,25\)
\(\frac{1}{5}x:\frac{2}{3}=3:\frac{1}{4}\)
\(\frac{1}{5}x:\frac{2}{3}=3.4\)
\(\frac{1}{5}x:\frac{2}{3}=12\)
\(\frac{1}{5}x=12.\frac{2}{3}\)
\(\frac{1}{5}x=8\)
\(x=8.5\)
\(x=40\)
a,3xy+x+15y+5
b,9-x^2-2xy-y^2
c,x^3-5x^2+x-5
d,x^2+y^2-2xy-1
tìm x
(5x-1)=(1-5x)^2
9-(x-2)^2=0
a. \(3xy+x+15y+15=x\left(3y+1\right)+15\left(y+1\right)=\left(x+15\right)\left(3y+1\right)\)
b.\(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3+x+y\right)\left(3-x-y\right)\)
c.\(x^3-5x^2+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x^2+1\right)\left(x-5\right)\)
d.\(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y+1\right)\left(x-y-1\right)\)
\(\left(5x-1\right)=\left(1-5x\right)^2\)
\(\left(5x-1\right)=\left(5x-1\right)^2\)
\(\left(5x-1\right)\left(1-5x+1\right)=0\)
\(\left(5x-1\right)\left(2-5x\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{2}{5}\end{array}\right.\)
\(9-\left(x-2\right)^2=0\)
\(\left(3+x-2\right)\left(3-x+2\right)=0\)
\(\left(1+x\right)\left(5-x\right)=0\)
\(\left[\begin{array}{nghiempt}x=-1\\x=5\end{array}\right.\)
a,3xy+x+15y+5
b,9-x^2-2xy-y^2
c,x^3-5x^2+x-5
d,x^2+y^2-2xy-1
tìm x
(5x-1)=(1-5x)^2
9-(x-2)^2=0
1 cho biểu thức A=5x(xy^2-2xy)-5x^2y^2. Rút gọn A .b) Tính GT của A khi x=-1/2 ,y=2
2. Tìm GTLN của bt A = |x-7|-|x-9|.Q= |x-2|+|x-8| b) tìm GTLN của bt P= 9-2|x-3|
Tìm x,y: (5x+1)^2=36/49=(2y-1)^3
(x-2/9)^3=(2/3)^6=(y/3)^2
(8x-1)^2y+1=5^2y+1 (với y thuộc N)
1) 3y/5=21/10
2) -11/12.y+0,25=6/5
3) 5x/4=-3/2
4) 8:(1/4.x)=2 : 0,02
bài 1 :tìm x,y biết
a) (5x+1)=\(\dfrac{36}{49}\) b) (x-2/9) = (2/3) c)(8x-1) 2x+1= 5^2 x+1
d) (x-3,5)^x+(y - 1/10)^4=0
`(5x+1)=36/49`
`<=> 5x = 36/49-1`
`<=> 5x = -13/49`.
`<=> x = -13/245.`
Vậy `x = -13/245`.
`b, x-2/9 = 2/3`.
`<=> x = 2/3 + 2/9`
`<=> x = 8/9`.
Vậy `x = 8/9`.
c: (8x-1)^(2x+1)=5^(2x+1)
=>8x-1=5
=>8x=6
=>x=3/4
d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0
=>x-3,5=0 và y-0,1=0
=>x=3,5 và y=0,1
Bài 1 : Tìm x,yez
1, 3 - | 2x + 1 | = ( - 5 )
2, 12 + | 3 - x | = 9
3 , | x + 9 | = 12 + ( - 9 ) + 2
4 , 5x - 16 = 40 + x
5 , 5x - 7 = - 21 - 2x
6, ( 2x - 1 ) ( y - 2 ) = 12
nhờ cấc a/c giúp e vs ạk e đang cần gấp
6, ( 2x + 1 ) ( y - 2 ) = 12
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
1.Tìm x biết
a, 12 . ( x - 1 ) = 0
b, 3/5x - 3/4 = ( - 3 )2 - | - 11|
c, 4 . ( x - 5/8 ) - 3/4 = 0,25
2. Tìm một số, biết 3/5 của nó bằng 27
a)12(x-1)=0
=>x-1=0
=>x=1
b) 3/5x-3/4=(-3)2-|-11|
3/5x-3/4=9-11=-2
3/5x=-2+3/4=-5/4
=>x=-5/4:3/5=-25/12
c)4(x-5/8)-3/4=0,25
4(x-5/8)=0,25+3/4=1
=>x-5/8=1/4
x=1/4+5/8=7/8
2) Số đó là
27:3/5=45