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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
Bài 1:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)
\(=4\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^4-3x-5x^3-x^2+x^2\)
\(=2x^4-5x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
Bài 14:Tìm x
a,\(x-3=\left(3-x\right)^2\)
\(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\)
\(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
b,\(\left(2x-5\right)-\left(5+2x\right)^2=0\)
\(\Rightarrow\left(2x-5\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)
a: \(A=4\cdot15^2-70^2=-4000\)
b: \(B=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(=100^2=10000\)
c: \(C=b^2-3b+a^2+3a-2ab\)
\(=\left(a-b\right)^2+3\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+3\right)\)
\(=\left(-5\right)\cdot\left(-5+3\right)=\left(-5\right)\cdot\left(-2\right)=10\)
d: \(D=\left(x-y\right)^3+3xy\left(x-y\right)+3xy\)
\(=\left(-1\right)^3-3xy+3xy\)
=-1
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Bài 1:
\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
Đẳng thức xảy ra khi \(a=b=1\)
Bài 2:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2=1^2=1\)
\(\Rightarrow x^2+y^2+z^2\ge\dfrac{1}{3}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)
Bài 3:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(4+1\right)\left(4x^2+y^2\right)\ge\left(4x+y\right)^2\)
\(\Rightarrow5\left(4x^2+y^2\right)\ge\left(4x+y\right)^2\)
\(\Rightarrow5\left(4x^2+y^2\right)\ge\left(4x+y\right)^2=1^2=1\)
\(\Rightarrow4x^2+y^2\ge\dfrac{1}{5}\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{5}\)
bài 1 mình thấy sao sao ý !!
đề bài là với mọi a,b,c tùy ý và chứng minh chứ bạn làm là khai thác ý cần chứng minh để chỉ ra điều kiện mà
a: \(A-B=\dfrac{\left(x-y\right)\left(x^2+y^2\right)-\left(x^2-y^2\right)\left(x+y\right)}{\left(x+y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x^3+xy^2-x^2y-y^3-x^3-x^2y+xy^2+y^3}{\left(x+y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{-2x^2y+2xy^2}{\left(x+y\right)\left(x^2+y^2\right)}=\dfrac{-2xy\left(x-y\right)}{\left(x+y\right)\left(x^2+y^2\right)}>0\)
=>A>B
b: \(A=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}< 3^{32}-1=B\)