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\(1,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8ay}{3}+16y^2\)
\(2,\)Bạn xem lại đề bài giùm mk nhé
\(\left(x^2+\frac{2}{5}y\right).\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)
yx=10⇒x=10y
M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}M=8x2−24xy16x2−40xy=8x(x−3y)8x(2x−5y)=x−3y2x−5y
=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}=10y−3y2.10y−5y=715
Câu 2
A) Với \(x>y>0\),ta có: \(x^2+y^2< x^2+y^2+2xy=\left(x+y\right)^2\Rightarrow\frac{1}{x^2+y^2}>\frac{1}{\left(x+y\right)^2}\)
Xét: \(\frac{x^2-y^2}{x^2+y^2}>\frac{x^2-y^2}{\left(x+y\right)^2}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x-y}{x+y}\)--->ĐPCM
B) \(3^{16}+1=\left(3^{16}-1\right)+2=\left(3^8+1\right)\left(3^8-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)+2\)
\(>\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)--->ĐPCM
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
Answer:
Câu 1:
\(\left(5x-x-\frac{1}{2}\right)2x\)
\(=\left(4x-\frac{1}{2}\right)2x\)
\(=4x.2x-\frac{1}{2}.2x\)
\(=8x^2-x\)
\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)
\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)
\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)
\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)
\(=x^4+8x^3+19x^2+24x+48\)
Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)
Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(= (x²+2xy+y²)-(x²-2xy+y²)\)
\(= x²+2xy+y²-x²+2xy-y²\)
\(= 4xy\)
\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)
Câu 2:
\(x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(x^2.\left(x-1\right)+4-4x=0\)
\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x-2=0\Rightarrow x=2\)
Trường hợp 3: \(x+2=0\Rightarrow x=-2\)
Câu 3: Bạn xem lại đề bài nhé.
1)
a) 25x^2-2xy+1/25y^2=(5x)^2-2.5.1/25xy+(1/5y)^2=(5x-1/5)^2
2)
a) (x+4)^2 -(x-1)(x+1)=16
x^2+8x+16-x^2+1=16
8x+17=16
x=-1/8