K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NN
16 tháng 7 2019
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 = \(\frac{3}{2}\)
HH
1 tháng 8 2018
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/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
23 tháng 9 2017
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:
\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =100.50=5000\)
\(b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\)
23 tháng 9 2017
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 !)
G
3
NC
15 tháng 8 2020
Bài làm:
a) \(\left(x+4\right)^2-1=0\)
\(\Leftrightarrow\left(x+4\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
b) \(\left(2x-3\right)^2=100\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=10\\2x-3=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=13\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{2}\\x=-\frac{7}{2}\end{cases}}\)
c) \(x^2+8x+16=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
d) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow x=\frac{3}{2}\)
1 tháng 7 2019
Bài 1 :
\(a,\)\(\left(x-4\right)^2-36=0\)\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)
\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)
\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)
\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)
\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\)\(\Leftrightarrow x=-4\)
\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)
LB
3
1 tháng 8 2018
\(\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy ...
\(4x^2-12x=-9\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+3^2=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy ...
\(\left(x+8\right)^2=121\)
\(\Rightarrow\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy ...
1 tháng 8 2018
a.(x-4)2 -36=0
⇔(x-4-6)(x-4+6)=0
⇔(x-10))(x+2)=0
✱x-10=0 => x=10
✱ x+2 =0 => x=-2
Vậy x=10 và x=-2
b) 4x2 -12 + 9 =0
⇔ (2x)2 -2.2x.3 + 32 = 0
⇔(2x-3)2 =0
⇔2x-3=0
⇔ x= \(\dfrac{3}{2}\)
c) (x+8)2 -121=0
⇔ (x+8)2 -112 =0
⇔ (x+8-11)(x+8+11) =0
⇔ (x-3) (x+19) =0
\(\begin{matrix}x-3=0\\x+19=0\end{matrix}\) ⇔ \(\begin{matrix}x=3\\x=-19\end{matrix}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 11 2019
ĐKXĐ; ...
a/ \(P=\frac{x^2}{x+4}\left[\frac{\left(x+4\right)^2}{x}\right]+9=x\left(x+4\right)+9=\left(x+2\right)^2+5\ge5\)
\(P_{min}=5\) khi \(x=-2\)
b/ \(Q=\left(\frac{\left(x+2\right)\left(x^2-2x+4\right).4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right).\frac{x\left(x-2\right)^3}{-16}\)
\(=\left(\frac{4\left(x^2-2x+4\right)-4x\left(x-2\right)}{\left(x-2\right)^2}\right).\frac{-x\left(x-2\right)^3}{16}\)
\(=\frac{16}{\left(x-2\right)^2}.\frac{-x\left(x-2\right)^3}{16}=-x\left(x-2\right)=-x^2+2x\)
\(=1-\left(x-1\right)^2\le1\)
\(Q_{max}=1\) khi \(x=1\)
Giải:
a) \(\left(x+8\right)^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-11\\x+8=11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-19\\x=3\end{matrix}\right.\)
Vậy ...
b) \(x^2+8x+16=0\)
\(\Leftrightarrow x^2+2.4.x+4^2=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
c) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.3+3^2=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a, (x+8)2=121
<=>\(\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy x\(\in\){3;-19}
b,x2+8x+16=0
<=>(x+4)2=0
<=> x+4=0
<=>x=-4
Vậy x=-4
c,4x2-12x=-9
<=> 4x2-12x+9=0
<=> (2x-3)2=0
<=> 2x-3=0
<=> 2x=3
<=> x=1,5
Vậy x=1,5