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
K
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KV
17 tháng 12 2023
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
17 tháng 12 2023
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
29 tháng 10 2021
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
YN
5 tháng 4 2022
`Answer:`
1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2.\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)
\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)
\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)
\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)
\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)
\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)
\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)
\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(v=x^2+=8x+11\)
Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)
\(=v^2-4^2+15\)
\(=v^2-1\)
\(=\left(v+1\right)\left(v-1\right)\)
\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
4) \(\left(x^2-a\right)^2-6x^2+4x+2a\)
\(=\left(x^2-a\right)\left(x^2-a\right)-6x^2+4x+2a\)
\(=\left(x^2-a\right).x^2-a\left(x^2-a\right)-6x^2+4x+2a\)
\(=x^4-ax^2-a.\left(x^2-a\right)-6x^2+4x+2a\)
\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)
\(=x^4-2ax^2+a^2-6x^2+2a+4x\)
6) \(a^2-b^2-c^2+2bc-2a+1\)
\(=\left(a^2-2a+1\right)-\left(b^2-2bc+c^2\right)\)
\(=\left(a-1\right)^2-\left(b-c\right)^2\)
\(=\left(a-b+c-1\right)\left(a+b-c-1\right)\)
7) \(4a^2-4b^2+16bc-16c^2\)
\(=4a^2-\left(4b^2-16bc+16c^2\right)\)
\(=\left(2a\right)^2-\left(2b-4c\right)^2\)
\(=\left(2a-2b+4c\right)\left(2a+2b-4c\right)\)
\(=2.\left(a-b-2c\right).2\left(a+b-2c\right)\)
\(=4\left(a-b-2c\right)\left(a+b-2c\right)\)
TH
0
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 = \(\frac{3}{2}\)