6x - ( - 5 ) = 17
( x + 2 ) (x -9 )
6x-(-5)=17
(x+2).(x-9)=0
6x-(-5)=17
6x+5=17
6x=17-5
6x=12
x=2
\(6x-\left(-5\right)=17\)
\(\Leftrightarrow6x=17+\left(-5\right)\)
\(\Leftrightarrow6x=12\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
\(\left(x+2\right)\left(x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=9\end{cases}}}\)
Vậy \(x\in\left\{-2;9\right\}\)
6x-(-5)=17
6x=17+(-5)
6x=12
x=12:6
x=2
(x+2).(x-9)=0
suy ra:x+2=0 hoặc x-9=0
x=0-2 x=0+9
x=-2 x=9
vậy x thuộc{-2,9}
Giải phương trình:
a) \(\sqrt{x^2+4}=\sqrt{2x+3}\)
b) \(\sqrt{x^2-6x+9}=2x-1\)
c) \(\sqrt{4x+12}=\sqrt{9x+17}-5\)
d) \(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
5) 6x – 5 = 3 + 4x 6) x – (17 – x) = x – 7
7) 9 – 25 = (7 – x) – ( 25 + 7) 8) 10 – 2.(4 – 3x) = - 4
9) (-12)2
.x = 56 + 10. 13x 10) 2x2
– 1 = 49
3. tìm x
a. 7x=-14
b. 6x-(-5) =17
c. (x+2). (x-9)=0
a) 7x = -14
=> x = -14/7
=> x = -2
b) 6x - (-5) = 17
=> 6x + 5 = 17
=> 6x = 17 +5
=> 6x = 22
=> x = 22/6 = 11/3
c) (x + 2) . (x-9) = 0
=> x + 2 = 0 hoặc x-9 = 0
=> x = -2 hoặc x=9
vậy x = -2; x= 9
a.7x=-14
\(\Leftrightarrow x=\dfrac{-14}{7}\)
\(\Leftrightarrow x=-2\)
Vậy x=-2.
b.6x-(-5)=17
\(\Leftrightarrow6x=17-5\)
\(\Leftrightarrow6x=12\)
\(\Leftrightarrow x=2\)
Vậy x=2
c.(x+2).(x-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=9\end{matrix}\right.\)
Vậy x=-2 hoặc x=9.
a) 7x=-14
x= -14 : 7
x= -2
b) 6x - (-5)= 17
6x + 5 =17
6x =17-5
6x = 12
x = 12:6
x = 2
c)(x+2). (x-9)=0
Vậy (x+2) hoặc (x-9) là 0
Vì ko có số tự nhiên nào +2=0
Nên 0 sẽ là x - 9
Vậy x=0+9 =9
Vây x=9
Bai1:Tim x thuoc Z biet:
a)7x=14
b)6x-(-5)=17
(x+2) (x-9)=0
a) 7x=14
x = 14 : 7
x = 2
b ) 6x - (-5) = 17
6x = 17+(-5)
6x = 12
x = 12 :6
x = 2
c) (x+2) ( x-9) = 0
=> x +2 = 0 hoặc x-9=0
=> x= -2 => x=9
mình làm rùi , đi
tìm x
(2-x)^3 +(3+x)(9-3x+x^2)+6x(1-x)=17
\(\left(2-x\right)^3+\left(3+x\right)\left(9-3x+x^2\right)+6x\left(1-x\right)=17\\ \Leftrightarrow8-3.2^2.x+3.2.x^2-x^3+3^3+x^3+6x-6x^2-17=0\\ \Leftrightarrow x^3-x^3+6x^2-6x^2-12x+6x=17-27-8\\ \Leftrightarrow-6x=-18\\ \Leftrightarrow x=\dfrac{-18}{-6}=3\\ Vậy:x=3\)
\(\sqrt{x^2+9}+\sqrt{x^2-2x+17}=x^2+6x-3\left(x>=0\right)\)
bạn ơi giúp mình đi
tìm x E Z , biết
a.7x= - 14 b. 6x-(-5)=17 c. (x+2)(x-9)=0
a,7x= -14
x= -14 /7
x= -2
b,6x-(-5)=17
6x=17+(-5)
6x=12
x=12/6
x=2
c,(x+2)(x-9)=o
suy ra(x+2)=0 hoặc (x-9)=0
nếu x+2=0 suy ra x= -2
nếu x-9=0 suy ra x=9
a) 7x=-14
x=-2
b) 6x-(-5) =17
6x . =12
x . =2
c) (x+2)(x-9)=0
=) * x+2=0=) x=-2
* x-9 =0=) x=9
-x + 6x = (-4)2:8 7+ 2 (5+x) = -9 -(x- 6)+2(9+x)=17 2 (x+5) - 4(6-x)= -68 |x|-58=9
-x + 6x = (-4)2:8
5x=16:8
5x=2
x=2/5
vậy x=2/5
7+ 2 (5+x) = -9
2(5+x)=-9-7
2(5+x)=-16
5+x=-16:2
5+x=-8
x=-8-5
x=-13
vậy x=-13
-(x- 6)+2(9+x)=17
-x+6+18+2x=17
-x+2x=17-6-18
x=-7
vậy x=-7
2 (x+5) - 4(6-x)= -68
2x+10-24-4x=-68
2x-4x=-68-10+24
-2x= -54
2x=54
x=54:2
x=27
vậy x=27
|x|-58=9
|x|=9+58
|x|=67
=>x=\(\pm\)67
vậy x=\(\pm\)67