{ [ ( 2x +14 ) .4-3 ] }-1 =0
Tìm x có giá trị nguyên để: (2x−3)(x−14)<0(2x−3)(x−14)<0\left(2x-3\right)\left(x-\dfrac{1}{4}\right)< 0
Tìm x
1) (2x-1)(x+3)(2-x)=0
2)x^3 + x^2 + x + 1 = 0
3) 2x(x-3)+5(x-3) =0
4)x(2x-7)-(4x-14)=0
5) 2x^3 + 3x^2 + 2x + 3 = 0
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
1, x(2x-7)-4x+14=0 2, x+x^2-x^3-x^4=0
3, 2x^3+3x^2+2x+3=0 4, 4x^2 - 25 - (2x-5)(2x+7)=0
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
x-3)^2-2(2x-7)(x-3)=0
( x-5 ) . ( 3 - x ) = 0
( 2x - 8 ) . ( 5-x ) =0
7x ( 2x -14 ) = 0
(2x-4) . ( 6-2x) =0
`#3107.\text {DN01012007}`
\(\left(x-5\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy, \(x\in\left\{3;5\right\}\)
_______
\(\left(2x-8\right)\cdot\left(5-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-8=0\\5-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\div2\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy, \(x\in\left\{4;5\right\}\)
_______
\(7x\left(2x-14\right)=0\\ \Rightarrow\left[{}\begin{matrix}7x=0\\2x-14=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x=14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=14\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy, \(x\in\left\{0;7\right\}\)
______
\(\left(2x-4\right)\cdot\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\6-2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\div2\\x=6\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy, \(x\in\left\{2;3\right\}.\)
Tìm x
(2x-7)+17=6
12-2.(3-3x)=-2
-14+3.(-x+5)=-20
-90:5.(-3-2x)=6
(x+1).(x-3)=0
(2x-2).(x+4)=0
(22+4).(x+3)=0
(5-x).(6-2x)=0
3.(x+1)+5=x+8
-4.(2x+9)-(-8x+3)-(x+13)=0
(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(\left(2x-7\right)+17=6\)
\(\left(2x-7\right)=6-17\)
\(2x-7=-11\)
\(2x=-11+7\)
\(2x=-4\)
\(x=-4:2\)
\(\Rightarrow x=-2\)
\(V\text{ậy x = -2}\)
Tìm x, biết:
1) 2x (x-3) + 5x - 15 = 0
2) x ( 2x - 7) - 4x + 14 = 0
3) x2 - 12x + 36 = 0
4) (x + 3) (x2 - 3x + 9) - x (x -1)(x + 1) -27 = 0
1) \(2x\left(x-3\right)+5x-15=0\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)
2) \(x\left(2x-7\right)-4x+14=0\)
\(x\left(2x-7\right)-2\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
3) \(x^2-12x+36=0\)
\(\left(x-6\right)^2=0\)
\(x-6=0\)
\(x=6\)
4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)
\(x^3+27-x^3+x-27=0\)
\(x=0\)
-3(2x-1)-5(3x+2)=14
-4(3x+5)+7(-2x-3)=-11
2(-3x+4)+3(x-5)=2(x+6)+1
-5(3x-1)-4(-2x+3)=0
-7(2x-1)+5(3x-4)=-3(x-5)
|x-3|=0
3|x+2|-5=16 cac ban lam nhanh nhe
mk giúp bạn câu cuối nhé:
3|x+2|-5=16
3|x+2|=16+5
3|X+2|=21
|x+2|=21:3
|x+2|=7
=>x+2=7 hoặc x+2=-7
+) với x+2=7 +) với x+2= -7
x=5. x=-9
vậy x€{5,-9}
nếu có TGian mk sẽ giải cho bạn mấy câu trên
cam ơn bạn nhé bạn có giup mình not câu trên trong vong ngay ko
3.(2x+1) -19=14
3.(2x+1)=14+19
3.(2x+1)=33
2x+1=33:3
2x+1=11
2.x=11-1
2.x =10
x=10:2
x=5
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2
Vậy phương trình có nghiệm là x = 2
a,2x(x-3)+5(x-3)=0
b,(x^2-4)+(x-2)(3-2x)=0
c,x^3-3x^2+3x-1=0
d,x(2x-7)-4x+14=0
e,(2x-5)^2-(x+2)^2=0
f,x^2-x-(3x-3)=0
Giups mình giải bài này nhé?