Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
ĐK:.......
Đặt $4x^2+4x+5=a\Rightarrow 8x^2+8x+11=2a+1; 4-4x^2-4x=9-a$
PT trở thành:
$\sqrt{a}+\sqrt{2a+1}=9-a\Leftrightarrow \sqrt{a}-2+\sqrt{2a+1}-3+(a-4)=0$
$\Leftrightarrow \frac{a-4}{\sqrt{a}+2}+\frac{2(a-4)}{\sqrt{2a+1}+3}+(a-4)=0$
$\Leftrightarrow (a-4)\left(\frac{1}{\sqrt{a}+2}+\frac{2}{\sqrt{2a+1}+3}+1\right)=0$
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ nên $a-4=0$
$\Rightarrow a=4$
$\Leftrightarrow 4x^2+4x+5=4$
$\Leftrightarrow 4x^2+4x+1=0\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=-\frac{1}{2}$
\(\sqrt{4x^2+4x+5}+\sqrt{8x^2+8x+11}=4-4x^2-4x\)
<=> \(\sqrt{\left(2x+1\right)^2+4}+\sqrt{2\left(2x+1\right)^2+9}=5-\left(2x+1\right)^2\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{\left(2x+1\right)^2+4}\ge2\\\sqrt{2\left(2x+1\right)^2+9}\ge3\end{matrix}\right.\)
=> VT \(\ge\) 5 mà VP \(\le\) 5
Mà VT = VP
=> 2x + 1 = 0
<=> x = \(\dfrac{-1}{2}\)
a)
\(5x\left(4-8x\right)+40\left(x^2-1\right)=3\\ \Leftrightarrow20x-40x^2+40x^2-40=3\\ \Leftrightarrow20x-40=3\\ \Leftrightarrow20\left(x-2\right)=3\\ \Leftrightarrow x-2=\frac{3}{20}\\ \Leftrightarrow x=\frac{43}{20}\)
b)
\(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\\ \Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\\ \Leftrightarrow80x-35=4\\ \Leftrightarrow80x=39\\ \Leftrightarrow x=\frac{39}{80}\)
a) \(5x\left(4-8x\right)+40\left(x^2-1\right)=3\)
\(\Leftrightarrow\)\(20x-40x^2+40x^2-40=3\)
\(\Leftrightarrow20x-40=3\)
\(\Leftrightarrow20x=43\)
\(\Leftrightarrow x=\frac{43}{20}\)
b, \(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\)
\(\Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\)
\(\Leftrightarrow80x-35=4\)
\(\Leftrightarrow80x=39\)
\(\Leftrightarrow x=\frac{39}{80}\)
a: =x^4-3x^5+4x^8
b: =2x^3+2x^2+4x
c: =4x^2+8x-5
d: =2x+3x^2+7x^4
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )
\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)
\(\Rightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow-3x+2x+x=10+4-14\)
\(\Leftrightarrow0=0\)
Vậy pt đã cho có nghiệm đúng với mọi x
1) \(4x^5y^2-8x^4y^2+4x^3y^2\)
\(=4x^3y^2\left(x^2-2x+1\right)\)
\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=4x^3y^2\left(x-1\right)^2\)
2) \(5x^4y^2-10x^3y^2+5x^2y^2\)
\(=5x^2y^2\left(x^2-2x+1\right)\)
\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=5x^2y^2\left(x-1\right)^2\)
3) \(12x^2-12xy+3y^2\)
\(=3\left(4x^2-4xy+y^2\right)\)
\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=3\left(2x-y\right)^2\)
4) \(8x^3-8x^2y+2xy^2\)
\(=2x\left(4x^2-4xy+y^2\right)\)
\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=2x\left(2x-y\right)^2\)
5) \(20x^4y^2-20x^3y^3+5x^2y^4\)
\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)
\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=5x^2y^2\left(2x-y\right)^2\)
1: 4x^5y^2-8x^4y^2+4x^3y^2
=4x^3y^2(x^2-2x+1)
=4x^3y^2(x-1)^2
2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)
3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)
4: \(=2x\left(4x^2-4xy+y^2\right)=2x\left(2x-y\right)^2\)
5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)
Trả lời:
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)
\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)
\(\Leftrightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow10-3x=10-3x\)
\(\Leftrightarrow-3x+3x=10-10\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy S = R với \(x\ne0;x\ne2\)
b: Ta có: \(G=-4x^2+8x-2003\)
\(=-4\left(x^2-2x+\dfrac{2003}{4}\right)\)
\(=-4\left(x^2-2x+1+\dfrac{1999}{4}\right)\)
\(=-4\left(x-1\right)^2-1999\le-1999\forall x\)
Dấu '=' xảy ra khi x=1
F=5-4x2+4
=(5+4)-4x2
= 9- 4x2
Vì 4x2 ≥0 => -4x2 ≤0 => 9-4x2 ≤9 ∀x
Dấu = xảy ra khi x=0
ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)
\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)
kết bạn ko^-^
= 8x ( 4x - 5 - 4x + 4 )
= 8x . ( -1 )
=-8x
học tốt