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1.
a) m > 2011
b) m<2011
c) m =2011
2.
a) \(m< \frac{-11}{20}\)
b)\(m>\frac{-11}{20}\)
3. -101 chia hết cho (a+7)
4. (3x-8) chia hết cho (x-5)
5. đề sai, N chứ ko phải n, tui ngu như con bòoooooooooooooooooooooo
5) Gọi \(d\inƯC\left(2m+9;14m+62\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}\Rightarrow\hept{\begin{cases}7\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}\left(14m+63\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}}\)
\(\Rightarrow\left(14m+63\right)-\left(14m+62\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=\left\{-1;1\right\}\)
\(\RightarrowƯC\left(2m+9;14m+62\right)=\left\{-1;1\right\}\)
Vậy \(x=\frac{2m+9}{14m+62}\)là p/s tối giản (Vì tử và mẫu của p/s có ƯC là 1)
Ta có: \(\frac{a^2+b^2+a+b}{ab}\) là số nguyên \(\Rightarrow\left(a^2+b^2+a+b\right)⋮d^2\)
Mà \(a^2,b^2⋮d^2\Rightarrow\left(a+b\right)⋮d^2\Rightarrow a+b\ge d^2\Rightarrow\sqrt{a+b}\ge d\) hay \(d\le\sqrt{a+b}\) (đpcm)
Câu hỏi của Nguyễn Thái Hà - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo nhé!
\(\left(2a-3\right)\left(\frac{3}{4}a+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2a-3=0\\\frac{3}{4}a+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2a=3\\\frac{3}{4}a=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a=\frac{3}{2}\\a=-\frac{4}{3}\end{array}\right.\)
\(\left(2a-3\right)\left(\frac{3}{4}a+1\right)=0\)
<=> \(\left[\begin{array}{nghiempt}2a-3=0\\\frac{3}{4}a+1=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}a=\frac{3}{2}\\a=-\frac{4}{3}\end{array}\right.\)
Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)
\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\)
\( \implies\) \(a+b+c=4x+4y+4z\)
\( \implies\) \(x+y+z=\frac{a+b+c}{4}\)
+)Ta có : \(a=2x+y+z\)
\(\iff\) \(a=x+\left(x+y+z\right)\)
\(\iff\) \(a-\left(x+y+z\right)=x\)
\(\iff\) \(a-\frac{a+b+c}{4}=x\)
\(\iff\) \(x=\frac{3a-b-c}{4}\)
+)Ta có :\(b=2y+z+x\)
\(\iff\) \(b=y+\left(y+z+x\right)\)
\(\iff\)\(b-\left(y+z+x\right)=y\)
\(\iff\) \(b-\frac{a+b+c}{4}=y\)
\(\iff\)\(y=\frac{3b-c-a}{4}\)
+)Ta có :\(c=2z+x+y\)
\(\iff\) \(c=z+\left(z+x+y\right)\)
\(\iff\) \(c-\left(z+x+y\right)=z\)
\(\iff\) \(c-\frac{a+b+c}{4}=z\)
\(\iff\)\(z=\frac{3c-a-b}{4}\)
\( \implies\) \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\)
\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)
\(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)
\(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)
Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được :
\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2
\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2
\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2
\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 6
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)
\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)
\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)
\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\)
Áp dụng TCDTSBN ta có :
\(\frac{a-b}{x}=\frac{b-c}{y}=\frac{a-c}{z}=\frac{\left(a-b\right)+\left(b-c\right)-\left(a-c\right)}{x+y-z}=\frac{0}{x+y-z}=0\)
\(\Rightarrow\frac{a-b}{x}=0\Rightarrow a-b=0\Rightarrow a=b\) (1)
\(\Rightarrow\frac{b-c}{y}=0\Rightarrow b-c=0\Rightarrow b=c\) (2)
\(\Rightarrow\frac{a-c}{z}=0\Rightarrow a-c=0\Rightarrow a=c\) (3)
Từ (1);(2) và (3) \(\Rightarrow a=b=c\) (đpcm)
\(a.\)Ta có:\(\frac{x}{y}+\frac{y}{x}\ge2\)
\(AM-GM:\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\left(đpcm\right)\)
\(b.\)Nếu x,y dương thì Áp dụng BĐT Cô-si ta có:\(\frac{3x}{y}+\frac{3y}{x}\ge2\sqrt{\frac{3x}{y}.\frac{3y}{x}}=6\)hay\(\frac{3x}{y}+\frac{3y}{x}\ge6\left(đpcm\right)\)
Nếu x,y âm ta có:\(\frac{3x}{y}+\frac{3y}{x}=\frac{3x^2}{xy}+\frac{3y^2}{xy}\ge2\sqrt{\frac{3x^2}{xy}.\frac{3y^2}{xy}}=6\left(đpcm\right)\)