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1/a) Ta có: \(A=x^4+\left(y-2\right)^2-8\ge-8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy GTNN của A = -8 khi x=0, y=2.
b) Ta có: \(B=|x-3|+|x-7|\)
\(=|x-3|+|7-x|\ge|x-3+7-x|=4\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le7\end{cases}}\Rightarrow3\le x\le7\)
Vậy GTNN của B = 4 khi \(3\le x\le7\)
2/ a) Ta có: \(xy+3x-7y=21\Rightarrow xy+3x-7y-21=0\)
\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\Rightarrow\left(x-7\right)\left(y+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\y=-3\end{cases}}\)
b) Ta có: \(\frac{x+3}{y+5}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất bằng nhau của dãy tỉ số, ta có:
\(\frac{x+3}{y+5}=\frac{3}{5}\Rightarrow\frac{x+3}{3}=\frac{y+5}{5}=\frac{x+y+8}{8}=\frac{16+8}{8}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+3}{3}=3\Rightarrow x+3=9\Rightarrow x=6\\\frac{y+5}{5}=3\Rightarrow y+5=15\Rightarrow y=10\end{cases}}\)
Bài 3: đề không rõ.
Bài 1:\(a,A=x^4+\left(y-2\right)^2-8\)
Có \(x^4\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow A\ge0+0-8=-8\)
Dấu "=" xảy ra khi \(MinA=-8\Leftrightarrow x=0;y=2\)
\(b,B=\left|x-3\right|+\left|x-7\right|\)
\(\Rightarrow B=\left|x-3\right|+\left|7-x\right|\)
\(\Rightarrow B\ge\left|x-3+7-x\right|\)
\(\Rightarrow B\ge\left|-10\right|=10\)
Dấu "=" xảy ra khi \(MinB=10\Leftrightarrow3\le x\le7\Rightarrow x\in\left(3;4;5;6;7\right)\)
P=a+{(a-3)-[(a+3)-(-a-2)]}
=a+a-3-a-3+a+2
=2a-4
Q=[a+(a+3)]-[(a+2)-(a-2)]
=a+a+3-a-2+a+2
=2a+3
=> P<Q
tk nha!
P=a+{(a-3)-[(a+3)-(-a-2)]}
=a+a-3-a-3+a+2
=2a-4
Q=[a+(a+3)]-[(a+2)-(a-2)]
=a+a+3-a-2+a+2
=2a+3
=> P<Q
\(\frac{3}{4}x-\frac{1}{2}=2\left(x-4\right)+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-\frac{1}{2}=2\text{x}-8+\frac{1}{4}x\)
\(\Leftrightarrow\frac{3}{4}x-2\text{x}-\frac{1}{4}x=-8+\frac{1}{2}\)
\(\Leftrightarrow\frac{3-8-1}{4}x=\frac{-15}{2}\)
\(\Leftrightarrow-\frac{3}{2}x=-\frac{15}{2}\Leftrightarrow x=\frac{-15}{-3}=5\)
Vậy x = 5
\(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\div\frac{2}{9}\)
\(\Rightarrow\left(x-1\right)=\frac{16}{9}\cdot\frac{9}{2}\)
\(\Rightarrow x-1=8\Rightarrow x=9\)
Vậy x = 9
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(\Rightarrow\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\div2\)
\(\Rightarrow\frac{x}{x+1}=\frac{2004}{2005}\)
\(\Rightarrow2005\text{x}=2004\left(x+1\right)\)
\(\Rightarrow2005\text{x}=2004\text{x}+2004\)
\(\Rightarrow2005\text{x}-2004\text{x}=2004\)
\(\Rightarrow x=2004\)
Vậy x = 2004