Cho a,b,c khác 0\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\), Tính giá trị biểu thức A= \(\frac{a^2+bc}{a^2+2bc}+\frac{b^2+ca}{b^2+2ca}+\frac{c^2+ab}{c^2+2ab}\)
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1) ta có \(\left(x+y\right)^2=x^2+2xy+y^2.\)
\(=\left(x^2+y^2\right)+2xy\)
\(=20+2.8\)(theo giả thiết x^2+y^2=20 , xy=8)
\(=36\)
Vậy với x^2+y^2=20, xy=8 thì (x+y)^2=36
2) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Rightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^2\right)^2-1^2\right]\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^4\right)^2-1^2\right]\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^8\right)^2-1^2\right]\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}\right)^2-1^2\)
\(\Leftrightarrow3M=2^{32}-1\)
\(\Rightarrow M=\frac{2^{32}-1}{3}\)
RÚT GỌN BIỂU THỨC N BẠN LÀM TƯƠNG TỰ NHA
\(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Rightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(...\)
\(...\)
Kết quả rút gọn \(N=\frac{7^{32}-1}{3}\)
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\(A=x^2-6x+11\)
\(A=\left(x^2-6x+9\right)+2\)
\(A=\left(x-3\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-3\right)^2=0\)
\(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Vậy GTNN của \(A\) là \(2\) khi \(x=3\)
\(B=x^2-20x+101\)
\(B=\left(x^2-20x+100\right)+1\)
\(B=\left(x-10\right)^2+1\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-10\right)^2=0\)
\(\Leftrightarrow\)\(x-10=0\)
\(\Leftrightarrow\)\(x=10\)
Vậy GTNN của \(B\) là \(1\) khi \(x=10\)
Chúc bạn học tốt ~
\(A=x^2-6x+11\)
\(A=\left(x^2-6x+9\right)+2\)
\(A=\left(x-3\right)^2+2\)
Mà \(\left(x-3\right)^2\ge0\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi : \(x-3=0\Leftrightarrow x=3\)
Vậy \(A_{Min}=2\Leftrightarrow x=3\)
b) \(B=x^2-20x+101\)
\(B=\left(x^2-20x+100\right)+1\)
\(B=\left(x-10\right)^2+1\)
Mà \(\left(x-10\right)^2\ge0\)
\(\Rightarrow B\ge1\)
Dấu "=" xảy ra khi : \(x-10=0\Leftrightarrow x=10\)
Vậy \(B_{Min}=1\Leftrightarrow x=10\)
c) \(C=x^2-4xy+5y^2+10x-22y+28\)
\(C=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)
\(C=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\)\(\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)
Mà \(\left(x-2y+5\right)^2\ge0\)
\(\left(y-1\right)^2\ge0\)
\(\Rightarrow C\ge2\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vây \(C_{Min}=2\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)
\(=\)\(\left(3x+2\right)\left[\left(3x\right)^2-3x.2+2^2\right]-\left(x^2-3^2\right)\)
\(=\)\(\left(3x\right)^3+2^3-x^2-3^2\)
\(=\)\(27x^3-x^2+8-9\)
\(=\)\(27x^3-x^2-1\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)^2+\left(x+2\right)^2-2x^2=3\)
\(x^2-6x+9+x^2+4x+4-2x^2=3\)
\(-2x+13=3\)
\(-2x=3-13\)
\(-2x=-10\)
\(x=\frac{-10}{-2}\)
\(x=5\)
Vậy \(x=5\)
\(x^2-2x+2\)
\(=\left(x^2-2x+1\right)+1\)
\(=\left(x-1\right)^2+1\)
a) cho x+y=1. Tính giá trị biểu thức x^3+ y^3+ 3xy
b) cho x-y=1. Tính giá trị biểu thức x^3- y^3- 3xy
![](https://rs.olm.vn/images/avt/0.png?1311)
x^3+ y^3+ 3xy
=(x+y)(x^2 -xy + y^2 ) + 3xy
=x^2 -xy + y^2 + 3xy
=x^2 + 2xy + y^2
=(x+y)^2 =1
=> x^3+ y^3+ 3xy=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+4x^3+12\)
\(=\left(x^2\right)^2+2.2x^3+\left(2x\right)^2-4x^2+12\)
\(=\left(x^2+2x\right)^2-4x^2+12\)
Có \(\left(x^2+2x\right)^2-4x^2+12>0\)
=> Vô nghiệm
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)