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\(a^2+2ab+b^2=\left(a+b\right)^2\ge0\forall a,b\)
\(a^2-2ab+b^2=\left(a-b\right)^2\ge0\forall a,b\)
\(A^{2n}\ge0\forall A\)
\(-A^{2n}\le0\forall A\)
\(\left|A\right|\ge0\forall A\)
\(-\left|A\right|\le0\forall A\)
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\left|A\right|-\left|B\right|\le\left|A-B\right|\)
\(A=\left|x+\dfrac{1}{5}\right|-x+\dfrac{4}{7}\)
Để A lớn nhất thì giá trị của x phải lớn nhất
\(\Leftrightarrow\)x là 1 số nguyên dương
Khi đó,
\(A=\left|x+\dfrac{1}{5}\right|-x+\dfrac{4}{7}=x+\dfrac{1}{5}-x+\dfrac{4}{7}\)
\(=\dfrac{1}{5}+\dfrac{4}{7}=\dfrac{27}{35}\)
Vậy \(A_{max}\)=\(\dfrac{27}{35}\)
\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
Giup minh voi minh can gap nha.Cai nao de lam truoc cung duoc. Cam on nhieu
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-x-3/5+4/5=-2x+12/35
b: B=|x-1/7|+|x+3/5|-1/3
x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
B=1/7-x+3/5+x-1/3=43/105
\(C=\dfrac{\left|x+5\right|+\left|7-x\right|+8}{\left|x+5\right|+\left|x-7\right|+3}\)
\(C=\dfrac{\left|x+5\right|+\left|7-x\right|+8}{\left|x+5\right|+\left|7-x\right|+3}\)
Đặt:
\(A=\left|x+5\right|+\left|7-x\right|\)
Áp dụng bđt:
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(A\ge\left|x+5+7-x\right|\)
\(A\ge12\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+5\ge0\Rightarrow x\ge-5\\7-x\ge0\Rightarrow x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x+5< 0\Rightarrow x< -5\\7-x< 0\Rightarrow x>7\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5\le x\le7\)
\(C\le\dfrac{12+8}{12+3}\)
\(C\le\dfrac{20}{15}\)
\(C\le\dfrac{4}{3}\)
Dấu "=" xảy ra khi:
\(-5\le x\le12\)