Tìm x, biết:
a) (x-4)2-36=0
b) 4x2-12x=-9
c) (x+8)2=121
Bài 1: Phân tích nhân tử
a. 16a^2 - 49( b - c)^2
b. (ax + by)^2 - (ax - by)^2
c. a^6 - 1
d. a^8 - b^8
Bài 2:Tìm x biết
a. (x - 4)^2 - 36 =0
b. (x + 8)^2 = 121
c. x^2 + 8x +16 =0
d. 4x^2 - 12x = -9
a. 16a2 - 49.( b - c )2
= ( 4a )2 - 72.( b - c )2
= ( 4a )2 - [ 7.( b - c ) ]2
= ( 4a )2 - ( 7b - 7c )2
= ( 4a - 7b + 7c ).( 4a + 7b - 7c )
b. ( ax + by )2 - ( ax - by )2
=( ax + by + ax - by ).( ax + by - ax + by )
= 2ax . 2by
= 2.( ax + by )
c.a6 - 1
= ( a3 )2 - 1
= ( a3 - 1 ).( a3 + 1 )
= ( a - 1 ).( a2 + a + 1 ).( a + 1 ).( a2 - a + 1 )
d. a8 - b8
= ( a4 )2 - ( b4 )2
= ( a4 - b4 ).( a4 + b4 )
= [ ( a2 )2 - ( b2 )2 ].( a4 + b4 )
= ( a2 - b2 ).( a2 + b2 ).( a4 + b4 )
= ( a - b ).( a + b ).( a2 + b2 ).( a4 + b4 )
B2
( x - 4 )2 - 36 = 0
\(\Leftrightarrow\) ( x - 4 )2 = 36
\(\Leftrightarrow\) ( x - 4 )2 = 62
\(\Leftrightarrow\) x + 4 = \(\pm\) 6
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
Vậy x = 10 , x = -2
b. ( x - 8 )2 = 121
\(\Leftrightarrow\) ( x - 8 )2 = 112
\(\Leftrightarrow\) x - 8 = \(\pm\)11
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=11\\x-8=-11\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=19\\x=-3\end{cases}}\)
Vậy x = 19 , x = -3
c. x2 + 8x + 16 = 0
\(\Leftrightarrow\)x2 + 2.4x + 42 = 0
\(\Leftrightarrow\) ( x + 4 )2 = 0
\(\Leftrightarrow\) x + 4 = 0
\(\Leftrightarrow\) x = -4
Vậy x = -4
d. 4x2 - 12x = - 9
\(\Leftrightarrow\)( 2x )2 - 2.2.x.3 + 32 = 0
\(\Leftrightarrow\) ( 2x - 3 )2 = 0
\(\Leftrightarrow\) 2x - 3 = 0
\(\Leftrightarrow\) 2x = 3
\(\Leftrightarrow\) \(x=\frac{3}{2}\)
Vậy x = \(\frac{3}{2}\)
Tìm x biết
a) \(\left(x-4\right)^2-36=0\)
b)\(\left(x+8\right)^2=121\)
c)\(x^2+8x+16=0\)
d)\(4x^2-12x+9=0\)
GIÚP MK VS MK ĐG CẦN GẤP AI NHANH MK TICK CHO
a/ \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
b/ \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)
<=> \(\left(x-3\right)\left(x+19\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)
d/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
\(c,x^2+8x+16=0\)
\(\Rightarrow x^2+4x+4x+16=0\)
\(\Rightarrow x.\left(x+4\right)+4.\left(x+4\right)=0\)
\(\Rightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x=-4\)
\(d,4x^2-12x+9=0\)
\(4x^2-6x-6x+9=0\)
\(\Rightarrow\left(2.2x^2-2.3x\right)-\left(3.2x-3^2\right)=0\)
\(\Rightarrow2.\left(2x^2-3x\right)-3.\left(2x-3\right)=0\)
\(\Rightarrow2x.\left(2x-3\right)-3.\left(2x-3\right)=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\Rightarrow x=1,5\)
Hai phần đầu dễ bãn tự làm
Bài 1:Tìm x
a) (x-4)^2 - 36 = 0
b) ( x-8)^2 = 121
c)x^2 + 8x + 16 = 0
d) 4x^2 - 12x = -9
Bài 2: Tính nhanh
a) 75^2 - 25^2
b) 53^2 - 47^2
Bài1:
\(â,\left(x-4\right)^2-36=0\\ \Leftrightarrow\left(x-4\right)^2=36\\ \Leftrightarrow x-4\in\left\{-6;6\right\}\\ \Leftrightarrow x\in\left\{-2;10\right\}\)
Vậy...
b<Tương tự
c,\(x^2+8x+16=0\\ \Leftrightarrow\left(x+4\right)^2=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)
Vậy...
d,Tương tự
Bài2:
\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =100.50=5000\)
\(b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\)
Bài 1:
a) (x-4)^2 - 36 = 0
=> (x-4)^2 = 36
=> (x-4)^2 = 6^2
=> x-4 = 6
=>x = 2
b) (x-8)^2 = 121
=> (x-8)^2 = 11^2
=> x-8 = 11
=> x = 19
c) x^2 + 8x +16 = 0
=> x( x +8) = -16
=> x = -4
d) 4x^2 - 12x = -9( Mk chưa nghĩ ra !)
Bài 2 :
a) 75^2 - 25^2
= (75-25)(75+25)
=50.100
=5000
b) 53^2 - 47^2
= (53-47)(53+47)
=6.100
=600
.Tìm x biết:
a) 3x(x – 2) – x + 2 = 0
b) x3 – 6x2 + 12x – 8 = 0
c) 16x2 – 9(x + 1)2
d) x2 (x – 1) – 4x2 + 8x – 4 = 0
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
TÌM X biết
a) ( x - 4 )2 - 36 = 0
b) 4x2 - 12x = -9
c) x( x - 5 ) - 4x + 20 = 0
a) (x - 4)2 - 36 = 0
=> (x - 4)2 = 36
=> x - 4 = 6 hoặc x - 4 = -6
=> x = 10 hoặc x = -2
b) hình như sai đề bn ạ
c) x(x - 5) - 4x + 20 = 0
=> x(x - 5) - 4(x - 5) = 0
=> (x - 5)(x - 4) = 0
=> x - 5 = 0 hoặc x - 4 = 0
=> x = 5 hoặc x = 4
1. tìm x
a) (x-4)^2 - 36=0
b) (x+8))^2=121
c) x^2 + 8x + 16=0
d) 4x^2 - 12x= -9
2 .CMR với mọi số nguyên n thì
a) (n+2)^2 - (n-2)^2 chia hết cho 8
b) (n+7)^2 - (n-5)^2 chia hết cho 24
MONG CÁC BẠN GIÚP NHANH CHO MK ĐỂ KỊP NỘP BÀI CHÂN THÀNH CẢM ƠN ^.^
Bài 1 :
\(a,\)\(\left(x-4\right)^2-36=0\)\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)
\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)
\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)
\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)
\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\)\(\Leftrightarrow x=-4\)
\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)
Bài 1 a) (x-4)^2 -36=0
=> (x-4)^2 = 36
=> x-4 = 6
=> x= 10
b) (x+8)^2 = 121
=> x+8 = 11
=> x=3
c) x^2 + 8x +16=0
=> (x+4)^2 =0
=> x+ 4 =0 => x= -4
d) 4x^2 - 12x= -9
=> 4x^2 -12x+9=0
=> ( 2x-3)^2=0
=> 2x-3 =0
=> x= 3/2
Bài 1)
a) (x+4)2 - 36 =0
(x+4)^2 - 6^2 =0
=> (x+2)(x-10) =0 (sử dụng hằng đẳng thức)
=> x =-2 ; x= 10(tự xét)
b) (x+8)^2 =121
(x+8)^2 - 121 =0
(x+8)^2 - 11^2 = 0
=> (x+19)(x-3) = 0
=> x= -19 ; x =3
c) x^2 + 8x + 16 =0
x^2 + 8x + 4^2 = 0
=> (x+4)^2 =0
=> x = -4
d) 4x^2 -12x =-9
(2x)^2 - 12x -9 = 0(chuyển vế)
=> (2x)^2 -12x -(3)^2=0
= (2x-3)^2 =0
=> x = 1,5
Bài 2:
a) (n+2)^2 - ( n-2)^2 = ( n +2+n-2)(n + 2 -n+2)
=> 2n * 4 = 8n : 8 với mọi n
b) (n+7)^2 - ( n-5)^2 = (n+7+n-5)(n+7-n+5)
= (2n +2)*12 = 24(n+1) : 24 với mọi x
Thực hiện phép tính:
a)-2x.(3x+4)+(2x+5).(9-7x)
b)(x-5)2-(4-x).(4+x)
c)(4x2-12x+8):(x-2)
1.rút gọn bt A= (x+2)3-2x(x+3)+(x3-8):(x-2)
2. tìm x biết:
a. 3x2-12x=0
b.4x2-1-4(1-2x)=0