x:3+3x-10=10
Những câu hỏi liên quan
c)3x^2-7x-10=0
d)2x(x-10)-x+10=0
e)3x^3+7x^2+17x+5=0
f)(2x-1)^2-(x-3)^2=0
g)x^3-5x^2+8x=4
c, \(3x^2-7x+10=0\)
\(\Leftrightarrow3x^2+3x-10x+10=0\)
\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)
Đúng 3
Bình luận (0)
d, \(2x\left(x-10\right)-x+10=0\)
\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)
Đúng 4
Bình luận (0)
Xem thêm câu trả lời
Tìm mẫu thức chung của hai phân thức\(\frac{x+1}{x^2+2x-3}\)và\(\frac{-2x}{x^2+7x+10}\)là:
A.\(x^3+6x^2+3x+10\)
B.\(x^3-6x^2+3x-10\)
C.\(x^3+6x^2-3x-10\)
D.\(x^3+6x^2+3x+10\)
Giải hộ mình vs
\(\text{A.}\)\(\text{x3+6x2+3x−10}\)
phân tích đa thức thành nhân tử
b)3x(x-2y)+4y(2y-x)+2(3x-4y)
f)1/3x(x-10-2/3x^2(x-10+3/2(x-1)x^3
h)8x(x-3y)+3y-x-8x+1
lẹ nha mn
Mày ra câu hỏi từ từ người ta trả lới cho chứ cứ hối người ta 😡
Đúng 1
Bình luận (0)
b) \(3x\left(x-2y\right)+4y\left(2y-x\right)+2\left(3-4y\right)\)
\(=3x\left(x-2y\right)-4y\left(x-2y\right)+2\left(3-4y\right)\)
\(=\left(x-2y\right)\left(3x-4y\right)+2\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left[\left(x-2y\right)+2\right]\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tìm x :
a. 10 x X - 1 - 3 - 5 - 7 - .... - 19 = 2 + 4 + 6 + .... + 20 .
b. 3x / 2 + 3x / 6 + 3x / 12 + 3x / 20 + 3x / 30 = 10
a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
Đúng 0
Bình luận (0)
a 10 x X-1-3-5-7-....-19 = 2+4+6+....+20
10xX-1-3-5-7-....-19=110
10xX=110+1+3+5+7+....+19
10xX=210
X=210:10
X=21
b là 4
Đúng 0
Bình luận (0)
b) x² + 1 -7√x²+1 +10=0 c) x²-3x -10 + 3√√x (x-3) = 0
Tìm x, biết:
a)x(2x-3)-(2x-1)(x+5)=17
b)(2x+5)^2+(3x-10)^2+2.(2x+5)(3x-10)=0
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
Đúng 2
Bình luận (0)
4(x-3)-8x(x-3)=0
5x(x-7)-10(7-x)=0
2x-8=3x(x-4)
3x(x-5)=10-2x
6x(x-3)-3(3-x)=0
x^2(x+4)+9(-x-4)=0
giup voi dang can gap a
\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)
Đúng 2
Bình luận (0)
\(\left(4-8x\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)
\(2\left(x-4\right)-3x\left(x-4\right)=0\)
\(\left(2-3x\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x+\sqrt{x^2+x-10}}{2x+3}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x^2+\sqrt{x^2+x-10}}{\sqrt{x^3+x^2-3x-x^2+3}}\)
28 tháng 4 2021 lúc 21:52
1. |-3x|= x+5
2. 10-|x+1|=3x+5
3.2x+3/-4 ≥ 4-x/-3
1: Ta có: |-3x|=x+5
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x+5\left(x\le0\right)\\3x=x+5\left(x>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-x=5\\3x-x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x=5\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\left(nhận\right)\\x=\dfrac{5}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{5}{4};\dfrac{5}{2}\right\}\)
Đúng 2
Bình luận (0)
2: Ta có: \(10-\left|x+1\right|=3x+5\)
\(\Leftrightarrow\left|x+1\right|=10-3x-5\)
\(\Leftrightarrow\left|x+1\right|=-3x+5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=-3x+5\left(x\ge-1\right)\\-x-1=-3x+5\left(x< -1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3x=5-1\\-x+3x=5+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=4\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy:S={1}
Đúng 2
Bình luận (0)
3: Ta có: \(\dfrac{2x+3}{-4}\ge\dfrac{4-x}{-3}\)
\(\Leftrightarrow\dfrac{-2x-3}{4}\ge\dfrac{x-4}{3}\)
Suy ra: \(3\left(-2x-3\right)\ge4\left(x-4\right)\)
\(\Leftrightarrow-6x-9-4x+16\ge0\)
\(\Leftrightarrow-10x\ge-7\)
hay \(x\le\dfrac{7}{10}\)
Vậy: S={x|\(x\le\dfrac{7}{10}\)}
Đúng 2
Bình luận (0)