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a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
1 M=\(x^2-4xy+4y^2-2x+4y+10\)
=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)
\(=\left(x-2y\right)\left(x-2y-2\right)+10\)
vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)
nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)
\(\Rightarrow\)A\(\ge13\)
dấu "=" xảy ra khi (x-2y)(x-2y-2)=0
\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)
vậy GTNN của M=10 khi x=0; y=0
x=2;y=1
\(Q\left(x\right)=x^2+2x-3=x^2+3x-x-3=\left(x+3\right)\left(x-1\right)\)
Q(x) có nghiệm\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
Áp dụng định lý Bezout:
\(P\left(x\right)⋮Q\left(x\right)\Leftrightarrow\hept{\begin{cases}P\left(-3\right)=0\\P\left(1\right)=0\end{cases}}\)
+) \(P\left(-3\right)=0\Leftrightarrow\left(-3\right)^4+3.\left(-3\right)^3-\left(-3\right)^2-3a+b=0\)
\(\Leftrightarrow81-81-9-3a+b=0\Leftrightarrow3a-b=-9\)(1)
+) \(P\left(1\right)=0\Leftrightarrow1^4+3.1^3-1^2+a+b=0\)
\(\Leftrightarrow1+3-1+a+b=0\Leftrightarrow a+b=-3\)(2)
Lấy (1) + (2), ta được:\(4a=-12\Leftrightarrow a=-3\)
Lúc đó \(b=-3+3=0\)
Vậy a = -3; b = 0
\(P\left(x\right)=x^4+3x^3-x^2+ax+b\)
\(Q\left(x\right)=x^2+2x-3\)
Để phép tính chia hết thì:
\(\Leftrightarrow\hept{\begin{cases}a+5=0\\b-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=-5\\b=3\end{cases}}}\)
Vậy ............
x2 - 5x = 0
=> x(x - 5) = 0
=> \(\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
b) (3x - 5)2 - 4 = 0
=> (3x - 5)2 = 0 + 4
=> (3x - 5)2 = 4
=> (3x - 5)2 = 22
=> \(\orbr{\begin{cases}3x-5=2\\3x-5=-2\end{cases}}\)
=> \(\orbr{\begin{cases}3x=7\\3x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)
Câu a):
ta có (x2-x-2)2+(x-2)2
=((x-2)2(x+1))2+(x-2)2
=(x-2)2(x2+2x+2)
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
a) 3x(x - 1) + 7x2(x - 1) = 0
<=> x(x - 1)(3 + 7x) = 0
<=> x = 0
hoặc : x - 1 = 0
hoặc 3 + 7x = 0
<=> x = 0
hoặc x = 1
hoặc x = -3/7
b) x2 - 2018x - 2019 = 0
<=> x2 - 2019x + x - 2019 = 0
<=> x(x - 2019) + (x - 2019) = 0
<=> (x + 1)(x - 2019) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)
c) (x + 3)2 - x(x - 2) = 13
<=> x2 + 6x + 9 - x2 + 2x = 13
<=> 8x = 13 - 9
<=> 8x = 6
<=> x= 6/8 = 3/4
a/\(3x\left(x-1\right)+7x^2\left(x-1\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(3x+7x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)x\left(3+7x\right)=0\)
Th1: x - 1 = 0
=> x = 1
Th2: x= 0
Th3: 3 + 7x = 0
=> x= -3/7
\(\Rightarrow x\in\left\{1;0;-\frac{3}{7}\right\}\)
b/ \(x^2-2018x-2019=0\)
\(\Leftrightarrow x^2+x-2019x-2019=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(2019x+2019\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-2019\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2019\right)\left(x+1\right)=0\)
Th1 : x -2019 = 0
=> x =2019
Th2: x + 1 =0
=> x = -1
\(\Rightarrow x\in\left\{2019;-1\right\}\)
c/ \(\left(x+3\right)^2-x\left(x-2\right)=13\)
\(\Leftrightarrow x^2+6x+9-x^2+2x=13\)
\(\Leftrightarrow8x=4\Rightarrow x=\frac{1}{2}\)