tìm x
2(x+4)+12=36
Bài 1. Tìm m để với mọi y>9 ta có m(căn y -3)(-4y)/(3-căn y) > y+1
Bài 2. Tìm m để phương trình x^2+4(m-1)x-12=0 có 2nghiệm pb x1, x2 thỏa mãn 4|x1-2|Căn (4-x2)=(x1+x2-x1x2-8)^2
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
72 + 36 x2 + 24 x 3 + 18 x 4 + 12 x 6 + 168
72 + 36 x 2 + 24 x 3 + 18 x 4 + 12 x 6 + 168
= 72 + 72 + 72 + 72 + 72 + 168
= 72 x 5 + 168
= 360 + 168
= 528
72 + 36 x2 + 24 x 3 + 18 x 4 + 12 x 6 + 168
tìm x biết :
918: 9 - . ( 72 - 36 x2 : 4 -12 ) =50
dấu chấm là x nha
Tìm x, biết:
x - 36 : 18 = 12
x - 36 : 18 = 12
=> x - 2 = 12
=> x = 2 + 12 = 14
x - 36 : 18 = 12
x - 2 = 12
x = 12 + 2
x = 14
Vậy x = 14
x-36:18=12
x-36=12 nhân 18
x-36=216
x=216+36
x=252
đề bài quy đồng
a, x+2/7x+42; -13x/x2-36
b, 7/4x+16; 15/x2-16
c, 12/x2-4; 2/x-3
d, 2x/x2-1; 5/2x+2
mọi ng giúp e với ạ
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
c, \(\dfrac{12}{x^2-4}\) = \(\dfrac{12}{\left(x-2\right).\left(x+2\right)}\) Đk \(x\) \(\ne\) \(\pm\) 2
\(\dfrac{2}{x-3}\) đk \(x\) \(\ne\) 3
\(\dfrac{12}{x^2-4}\) = \(\dfrac{12.\left(x-3\right)}{\left(x^2-4\right).\left(x-3\right)}\) = \(\dfrac{12x-36}{\left(x^2-4\right).\left(x-3\right)}\)
\(\dfrac{2}{x-3}\) = \(\dfrac{2.\left(x^2-4\right)}{\left(x-3\right).\left(x^2-4\right)}\)
tìm x : 5/6-(x2/8)=1/4
(x-3/5)×3/4=21/20
49/45:x/9=7/5
x×1/5+x×1/5+x×3/5=8/9
8/9:x-5/9:x=1/36
5/4×(x×2)=11/4
a) 2x2 + 2x(5 - x)=12 d) 2(x + 5) - x2 - 5x = 0 g) (3x + 1)2 - (x+1) = 0
b) (5 - 2x)2 - 16 = 0 e) (2x - 1)2 - 4(x + 7)(x - 7) = 0 h) x2 + 7x - 8 = 0
c) 3x2 - 3x(x-2) = 36 f) (x + 4)2 - (x + 1)(x - 1) = 16 i) -2x2 +13x -15 = 0
mik cần gấp, cảm ơn mọi người.
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
tìm x,y biết:\(\frac{x}{12}=\frac{y}{3}\) x - y = 36
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{12}\)=\(\frac{y}{3}\)=\(\frac{x-y}{12-3}\)=\(\frac{36}{9}\)=4
=> x= 4.12= 48
y= 4.3= 12
Vậy .......
\(\frac{x}{12}-\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)
\(\Rightarrow\) TSP: 4
-) x = 4 . 12 = 48
-) y = 4 . 3 = 12
Vậy x = 48 ; y = 12
Chúc bạn học tốt!