Tìm x, biết:
a) (x-4)2-36=0
b) 4x2-12x=-9
c) (x+8)2=121
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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 !)
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
13 tháng 12 2021
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
19 tháng 8 2016
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 tháng 7 2019
Bài 1 :
\(a,\)\(\left(x-4\right)^2-36=0\)\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)
\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)
\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)
\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)
\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)
\(\Rightarrow x+4=0\)\(\Leftrightarrow x=-4\)
\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)
\(\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
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 ...
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
\(\begin{matrix}x-3=0\\x+19=0\end{matrix}\) ⇔ \(\begin{matrix}x=3\\x=-19\end{matrix}\)