Tim so nguyen x
a,3x :x-2
b,4x-5:x+1
Những câu hỏi liên quan
a) tim so nguyen x sao cho : 4x+5 chia het cho 2x-1
b) tim so nguyen x va y biet : xy+3x-2y -11 = 0
giup mik nha ! ai nhanh nhat mik tick cho !
tim so nguyen x: (12x-1)(6x-1)(4x-1)(3x-1)=330
cách 1: phân tích ra ước
cách 2 áp dụng 7 hằng đẳng thức nhân tung ra
Đúng 0
Bình luận (0)
ko f l tfboys mà là TFBOYS nhé , bn có f Tứ Diệp Thảo ko vx
Đúng 0
Bình luận (0)
tim so nguyen x bt
25 – 2x = 16 – 3x
d) x – (47 – 22) = 5+ (10 – 4x) ai giup mik vs
25 – 2x = 16 – 3x
-2x + 3x = 16 - 25
x = -9
Vậy x = -9
x - (47 - 22 ) = 5 + (10-4x)
x - 47 + 22 = 5 + 10 - 4x
x + 4x = 5 + 10 - 22 + 47
5x = 40
x = 40 : 5
x = 8
Vậy x = 8.
~ HOK TỐT ~
\(\text{25 – 2x = 16 – 3x}\)
\(2x+3x=16-25\)
\(5x=-9\)
\(\Rightarrow x=\frac{-9}{5}\)
\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)
\(x-47+22=5+10-4x\)
\(x+4x=5+10-22+47\)
\(5x=40\)
\(\Rightarrow x=8\)
học tốt
\(25-2x=16-3x\)
\(\Leftrightarrow25-2x-16=-3x\)
\(\Leftrightarrow-2x+9=-3x\)
\(\Leftrightarrow-3x+2x=9\)
\(\Leftrightarrow-x=9\Leftrightarrow x=9\)
Vậy ...
\(x-\left(47-22\right)=5+\left(10-4x\right)\)
\(\Leftrightarrow x-47+22=5+10-4x\)
\(\Leftrightarrow x-25=5+10-4x\)
\(\Leftrightarrow x-25=-4x+15\)
\(\Leftrightarrow x=-4x+15+25\)
\(\Leftrightarrow x=-4x+40\)
\(\Leftrightarrow x+4x=40\Leftrightarrow5x=40\Leftrightarrow x=8\)
Vậy ...
Xem thêm câu trả lời
tim gia tri x nguyen de bieu thuc (3x2-4x+x-1)/(3x+2) co gia tri nguyen
tim gia tri x nguyen de bieu thuc (3x3-4x2+x+1)/(x-4) co gia tri nguyen
tim so nguyen x bt
x – 18 = 22 – 47
b) 4 2 – x = 53 – 60
c) 25 – 2x = 16 – 3x
d) x – (47 – 22) = 5+ (10 – 4x)
a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
so cac so nguyen x thoa man 15-|-2x+3|*|5+4x|=19 tim so cac so nguyen x giup minh nhe
!
Tim x nguyen de phan so sau la so nguyen
4x-1/3-x
tim x
a)4x(x-7)-4x2=56
b)12x(3x-2)-(4-6x)=0
c)4(x-5)-(5-x)2=0
a: Ta có: \(4x\left(x-7\right)-4x^2=56\)
\(\Leftrightarrow4x^2-7x-4x^2=56\)
hay x=-8
b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)
\(\Leftrightarrow36x^2-24x-4+6x=0\)
\(\Leftrightarrow36x^2-18x-4=0\)
\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
Đúng 1
Bình luận (0)
tim x
a) (5x-3)^2-(4x-7)^2
b) (2x+7)^2=9(x+2)^2
c) (x+2)^2=9(x^2-4x+4)
b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Đúng 1
Bình luận (0)