\(x^2-4x+4=25\)
Giải phương trình
a/ 2x - (x - 3)(5 - x) = (x+4)\(^2\)
b/ (4x + 1)(x - 2) + 25 = (2x+3)\(^2\) - 4x
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
\(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}\)
Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.
* Coi đây là bài toán rút gọn
Lời giải:
ĐKXĐ: $x\neq 2$
$\frac{x-5}{x^2-4x+4}:\frac{x^2-25}{2x-4}=\frac{x-5}{(x-2)^2}:\frac{(x-5)(x+5)}{2(x-2)}$
$=\frac{x-5}{(x-2)^2}.\frac{2(x-2)}{(x-5)(x+5)}=\frac{2}{(x-2)(x+5)}$
Bài 1 khai triển các hằng đẳng thức
K) 4.x^2=
L) 1/9x^2 -25/16 y ^2
M)1/4x^2-4x^2
N) 4 /49 -4x^2
O) (x-3) (x+3)
P (x +4) (x -4 )
Mik cần gấp giúp mik vs
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
Phân tích đa thức thành nhân tử
4x^4+4x^2+1
9x^4-6x^+1
\(\dfrac{x^2}{9}\)-\(\dfrac{2}{3}\)x+1
x^2-25
\(4x^4+4x^2+1=\left(2x^2+1\right)^2\)
\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)
\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)
\(x^2-25=\left(x-5\right)\left(x+5\right)\)
a,(x+3)(x^2-3x+9)-(x-3)(x^2+3x+9)
b,(x-5)(x^2+5x+25)-(x+5)(x^2-5x+25)
c,(x-4)(x^2+4x+16)-(x+4)(x^2-4x+18)
d,(x-2)(x^2+7x+49)-(x-7)(x^2-7x+49)
a) = (x+3).(x-3)^2-(x-3)(x+3)^2
=(x^2-9)(x-3)-(x^2-9)(x+3)
=(x^2-9)(x-3-x-3)
=-6(x^2-9)
các câu còn lại tương tự
\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+3-\left(x^3-3\right)\)
\(=x^3+3-x^3+3\)
\(=6\)
\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)
\(=x^3-5^3-x^3-5^3\)
\(=-125-125\)
\(=-250\)
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(x+\sqrt{5-4x}=0\)
\(\sqrt{1-2x^2}=x-1\)
a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow\sqrt{x-1}=1\)
hay x=2
c: Ta có: \(\sqrt{1-2x^2}=x-1\)
\(\Leftrightarrow1-2x^2=x^2-2x+1\)
\(\Leftrightarrow-3x^2+2x=0\)
\(\Leftrightarrow-x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
7hằng đẳng thức.
x2 + 4x + 4=
4x2 - 4x + 1=
4x2 + 12x +9=
9x2 + 30x +25=
4x2 - 20x + 25 =
a. \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b. \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c. \(4x^2+12x+9=\left(2x\right)^2+2\cdot2x\cdot3+3^2=\left(2x+3\right)^2\)
d. \(9x^2+30x+25=\left(3x\right)^2+2\cdot3x\cdot5+5^2=\left(3x+5\right)^2\)
e. \(4x^2-20x+25=\left(2x\right)^2-2\cdot2x\cdot5+5^2=\left(2x+5\right)^2\)
\(x^2+4x+4=\left(x+2\right)^2\)
\(4x^2-4x+1=\left(2x-1\right)^2\)
\(4x^2+12x+9=\left(2x+3\right)^2\)
\(9x^2+30x+25=\left(3x+5\right)^2\)
\(4x^2-20x+25=\left(2x+5\right)^2\)
tik mik nha
Bài tập:Giải các phương trình sau
1)\(\sqrt{-4^2+25}=x\)
2)\(\sqrt{x^2-10x+25}\)=2x+1
3)\(\sqrt{x^2-6x+9}+x=11\)
4)\(\sqrt{x^2-4x+3}=x-2\)
Tính:
a,12xy-4x^2-9y^2
b,10x-x^2-25
c,x^4+4-4x^2
d,1/6 x^2-x+4
a) \(12xy-4x^2-9y^2=-\left(4x^2-12xy+9y^2\right)\)
= \(-\left(2x-3y\right)^2\)
b) \(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
c) \(x^4+4-4x^2\)
= \(\left(x^2-2\right)^2\)
Giải các PT sau bằng cách đưa về dạng PT tích
b) x^2+10x+25-4x.(x+5)=0 c/(4x-5)^2-2.(16x^2-25)=0 d/(4x+3)^2=4.(x^2-2x+1) e/ x^2-11x+28=0
x2+10x+25-4x(x+5)=0
⇔(x+5)2-4x(x+5)=0
⇔(x+5)(x+5-4x)=0
⇔(x+5)(5-3x)=0
⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)