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\(x^2-5x+6=0\)
=>\(x^2-2x-3x+6=0\)
=>x(x-2)-3(x-2)=0
=>(x-2)(x-3)=0
=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)
Xơi luôn nha:v
Có: \(\left(a^2+b^2+c^2\right)\left(5^2+1^2+1^2\right)\ge\left(5a+b+c\right)^2\)
Do đó \(A\ge\frac{\left(5a+b+c\right)^2}{27}\). Lại có: \(5a+b+c=4a+\left(a+b+c\right)\ge4.5+7=27\)
Từ đó \(A\ge27\)
True?
Từ \(a\ge5\)và \(a+b\ge6\)\(\Rightarrow b\ge1\)
Từ \(a+b\ge6\)và \(a+b+c\ge7\)\(\Rightarrow c\ge1\)
\(\Rightarrow A=a^2+b^2+c^2\ge5^2+1^2+1^2=27\)
Dấu = xảy ra khi \(a=5,b=c=1\)
Vậy \(minA=27\Leftrightarrow a=5,b=c=1\)
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.
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm5\end{cases}}\)
\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)
\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{\left(x^2+10x+25\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)x}\)
\(\Leftrightarrow M=\frac{\left(x+5\right)^2}{x\left(x+5\right)}\)
\(\Leftrightarrow M=\frac{x+5}{x}\)
b) Để \(M\inℤ\)
\(\Leftrightarrow x+5⋮x\)
\(\Leftrightarrow5⋮x\)
\(\Leftrightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Mà \(x\ne\pm5\)
\(\Leftrightarrow x\in\left\{1;-1\right\}\)
Vậy để \(M\inℤ\Leftrightarrow x\in\left\{1;-1\right\}\)
\(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\left(x\ne\pm5;x\ne0\right)\)
\(\Leftrightarrow M=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\left(\frac{x^2-5x}{\left(x-5\right)\left(x+5\right)}+\frac{5x+25}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}\cdot\frac{x-5}{x}\)
\(\Leftrightarrow M=\frac{\left(x+5\right)^2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)x}=\frac{x+5}{x}\)
b) M là số nguyên thì x+5 chia hết cho x
=> 5 chia hết cho x
x nguyên => x thuộc Ư (5)={-5;-1;1;5}
Vậy x={-5;-1;1;5} thì M là số nguyên
Chứng minh BĐT Cauchy-schwarz:
Xem câu hỏi
Áp dụng BĐT Cauchy-schwarz ta có:
\(P=a^2+2b^2+3c^2=a^2+\frac{b^2}{\frac{1}{2}}+\frac{c^2}{\frac{1}{3}}\ge\frac{\left(a+b+c\right)^2}{1+\frac{1}{2}+\frac{1}{3}}=\frac{1}{\frac{11}{6}}=\frac{6}{11}\)
Dấu " = " xảy ra \(\Leftrightarrow a=2b=3c\)
\(\Leftrightarrow b=\frac{3}{2}c\)
Có: \(a+b+c=1\)
\(\Leftrightarrow3c+\frac{3}{2}c+c=1\)
\(\Leftrightarrow\frac{11}{2}c=1\Leftrightarrow c=\frac{2}{11}\)
\(\Leftrightarrow\hept{\begin{cases}a=3c=\frac{6}{11}\\b=\frac{3}{2}c=\frac{3}{11}\end{cases}}\)
Vậy \(P_{min}=\frac{6}{11}\Leftrightarrow\hept{\begin{cases}a=\frac{6}{11}\\b=\frac{3}{11}\\c=\frac{2}{11}\end{cases}}\)
Thử cách này có phải ý bạn không:
\(P=\left(a^2+\frac{36}{121}\right)+\left(2b^2+\frac{18}{121}\right)+\left(3c^2+\frac{12}{121}\right)-\frac{6}{11}\)
\(\ge2\sqrt{a^2.\frac{36}{121}}+2\sqrt{2b^2.\frac{18}{121}}+2\sqrt{3c^2.\frac{12}{121}}-\frac{6}{11}\)
\(=\frac{12\left(a+b+c\right)}{11}-\frac{6}{11}=\frac{12}{11}-\frac{6}{11}=\frac{6}{11}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a^2=\frac{36}{121}\\2b^2=\frac{18}{121}\\3c^2=\frac{12}{121}\end{cases}}\) và a,b,c > 0 tức là \(\hept{\begin{cases}a=\frac{6}{11}\\b=\frac{3}{11}\\c=\frac{2}{11}\end{cases}}\) (t/m)
Vậy \(P_{min}=\frac{6}{11}\Leftrightarrow\)\(\hept{\begin{cases}a=\frac{6}{11}\\b=\frac{3}{11}\\c=\frac{2}{11}\end{cases}}\)