(3x - 10):10=20
Vậy x=?
|x +10| - (5 - 3x) = (4x - 10) - (x - 5)
|3x + 21| - ( 10 - 5x ) = 5x - |-20|
a) | x + 10 | - ( 5 - 3x ) = ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = ( 5 - 3x ) + ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = 5 - 3x + 4x - 10 - x + 5
=> | x +10 | = 0
=> x + 10 = 0
=> x = -10
Vậy...
b) Làm tương tự
Kết quả : | 3x + 21 | = -10 ( vô lí) ( vì |3x+21| >= 0 mà -10<0)
Vậy không tìm được x thỏa mãn bài toán
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
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
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, \(3x^2-7x+10=0\)
\(\Leftrightarrow3x^2+3x-10x+10=0\)
\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)
d, \(2x\left(x-10\right)-x+10=0\)
\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)
I: thu gọn
M= (-x-8)-(-3x+10)-(x-10)
N=-(x-100)+(-3x+10)-(-x-100)
Q=100-(-4x+1)-(99+x)-(x-1)
Help me
\(M=\left(-x-8\right)-\left(-3x+10\right)-\left(x-10\right)\\ =-x-8+3x-10-x+10\\ =\left(-x+3x-x\right)+\left(-8-10+10\right)\\ =x-8\)
\(N=-\left(x-100\right)+\left(-3x+10\right)-\left(-x-100\right)\\ =-x+100+-3x+10+x+100\\ =\left(-x+-3x+x\right)+\left(100+10+100\right)\\ =-3x+210\\ =3\left(-x+70\right)\)
\(Q=100-\left(-4x+1\right)-\left(99+x\right)-\left(x-1\right)\\ =100+4x-1-99-x-x+1\\ =\left(4x-x-x\right)+\left(100-1-99+1\right)\\ =2x+1\)
3x(x-10)=x-10
\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)
\(\Rightarrow\left(x-10\right)\left(3x-1\right)=0\)
\(\Rightarrow x-10=0\Rightarrow x=10\)
hoặc \(3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
vậy x = 10 ; x = 1/3
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
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 😡
b) \(3x\left(x-2y\right)+4y\left(2y-x\right)+2\left(3-4y\right)\)
\(=3x\left(x-2y\right)-4y\left(x-2y\right)+2\left(3-4y\right)\)
\(=\left(x-2y\right)\left(3x-4y\right)+2\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left[\left(x-2y\right)+2\right]\)
p(x)=(x^2-9x-10)^10+(3x^10+5x^7+2)^20=0
cần giúp gấp
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
\(\text{A.}\)\(\text{x3+6x2+3x−10}\)
x:3+3x-10=10
\(\frac{x}{3}+3x-10=10\)
\(\frac{x}{3}+\frac{9x}{3}=10+10=20\)
\(\frac{10x}{3}=20\)
\(10x=20.3=60\)
\(x=\frac{60}{10}=6\)