Tìm x biết: (x-3)^2-16=0
x^2 -2x=24
(2x+1)+(x+3)^2(x+7)(x-7)=0
Tìm x, biết:
a, (x-3)^2-4=0
b, x^2-2x=24
c,(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
b/ x2-2x=24
=> x2-2x-24=0
=> (x-6)(x+4)=0
=>x=6 hoặc x =-4
a/ (x-3)2 - 4 = 0
=> (x-3-2)(x-3+2)=0
=> (x-5)(x-1)=0
=> x = 5 hoặc x=1
Tìm x ,biết :
a, x^2 -2x =24
b, (2x-1)^2 + (x+3 )^2 -5 .(x+7) (x-7) =0
\(a,x^2-2x=24\)
\(x^2-2x-24=0\)
\(x^2-2x+1-25=0\)
\(\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(2x+255=0\)
\(2x=-255\)
\(x=-\frac{255}{2}\)
a/ \(x^2-2x=24\)
<=> \(x^2-2x+1-1=24\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=25\\x-1=-25\end{cases}}\)<=> \(\orbr{\begin{cases}x=26\\x=-24\end{cases}}\)
b/ \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
<=> \(2x+255=0\)
<=> \(2x=-255\)
<=> \(x=-\frac{255}{2}\)
bài 9:tìm x
1) (x-3)^2-4=0
2) x^2-2x=24
3) (2x-1)^2+(x+3)^3-5(x+7)(x-7)=0
giúp mình với mn ơi
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Tìm x biết:
a:(x-1)^3-x(x-2)^2-(x-2)=0
b:(2x+5)(2x-7)-(4x+3)^2=16
a) (x - 1)3 - x(x - 2)2 - (x - 2) = 0
<=> x3 - 2x2 + x - x2 + 2x - 1 - x3 + 4x2 - 4x - x + 2 = 0
<=> x2 - 2x + 1 = 0
<=> x2 - 2.x.1 + 12 = 0
<=> (x - 1)2 = 0
x - 1 = 0
x = 0 + 1
x = 1
=> x = 1
a)Ta có : \(\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)
\(=>\left(x-1\right)^3-\left(x^2-2x\right)\left(x-2\right)-\left(x-2\right)=0\)
\(=>\left(x-1\right)^3-\left(x-2\right)\left(x^2-2x+1\right)=0\)
\(=>\left(x-1\right)^3-\left(x-2\right)\left(x-1\right)^2=0\)
\(=>\left(x-1\right)^2\left(x-1-x+2\right)=0\)
\(=>\left(x-1\right)^2=0=>x-1=0=>x=1\)
Vậy x=1
b)(2x+5)(2x-7)-(4x+3)2=16
\(=>4x^2-4x-35-16x^2-24x-9-16=0\)
\(=>-\left(12x^2+28x+60\right)=0\)
\(=>12\left(x^2+\frac{7}{3}x+\frac{5}{3}\right)=0\)
\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)
Lại có \(\left(x+\frac{7}{6}\right)^2\ge0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}>0\)
Vậy ko có giá trị nào của x thỏa mãn đề bài
\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)
tìm x , biết
a) 5/6 - x = -7 / 12 + 2/3
b) ( 2,4x - 36 ) : 1va5/7 = -1
c) 2 ( x - 1 ) - 3 ( 2x + 2 ) - 4 ( 2x + 3 ) = 16
d ) ( x - 2 ) . ( x + 3 ) > 0
tìm x
(x+4)2- (x+1)(x-1)= 16
(2x-1)2+ (x+3)2- 5(x+7)(x-7)=0
a) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
(x + 4)2 - (x + 1) (x - 1) = 16
<=> (x2 + 8x + 16) - (x2 - 1) = 16
<=> x2 + 8x + 16 - x2 + 1 = 16
<=> 8x + 17 = 16
<=> 8x = -1
<=> x =
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2+1-4x+\left(x^2+9+6x\right)-5\left(x^2-7^2\right)=0\)
\(4x^2+1-4x+x^2+9+6x-5x^2+245=0\)
\(\left(4x^2+x^2-5x^2\right)-\left(4x+6x\right)+\left(1+9+245\right)=0\)
\(2x+255=0\)
\(2x=-255\)
\(x=\dfrac{-255}{2}\)
P/s: Nhớ tick cho mình nha. Thanks bạn
a) Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
hay \(x=-\dfrac{1}{8}\)
b) Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)
\(\Leftrightarrow2x=-255\)
hay \(x=-\dfrac{255}{2}\)
Tìm x, biết:
a) (2x+2)(x-1)-(x+2)(2x+1)=0;
b)(3x+1)(2x-3)-6x(x+2)=16;
c)(12x-5)(4x-1)+(3x-7)(1-16x)=81
mn ơi giúp mik vs ạ :<
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
Tìm x
(x-3)2-4=0
x2-2x=24
(2x-1)2+(x+3)2-5(x+7)(x-7)=0
Tìm x, biết :
a) (x+4)^2 - (x+1)×(x-1)=16
b) (2x-1)^2+(x-3)^2-5×(x+7)×(x-7)=0
(x + 4)2 - (x + 1)(x - 1) = 16
=> x2 + 8x + 16 - x2 + 1 = 16
=> 8x + 17 = 16
=> 8x = -1
=> x = -1/8
b) (2x - 1)2 + (x - 3)2 - 5(x + 7)(x - 7) = 0
=> (4x2 - 4x + 1) + (x2 - 6x + 9) - 5(x2 - 49) = 0
=> 5x2 - 10x + 10 - 5x2 + 245 = 0
=> -10x + 255 = 0
=> 10x = 255
=> x = 25,5
Vậy x = 25,5