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Ta có : \(\hept{\begin{cases}\left|5-\frac{2}{3}x\right|\ge0\forall x\\\left|\frac{1}{7}y-3\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5-\frac{2}{3}x\right|+\left|\frac{1}{7}y-3\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}5-\frac{2}{3}x=0\\\frac{1}{7}y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\y=21\end{cases}}\)
b) Ta có \(\hept{\begin{cases}\left|5x+10\right|\ge0\forall x\\\left|6y-9\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5x+10\right|+\left|6y-9\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}5x+10=0\\6y-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1,5\end{cases}}\)
\(a,\frac{-9}{x}=\frac{-9}{\frac{4}{49}}\)
\(\Rightarrow x=\frac{4}{49}\)
\(b,\left|x-2\right|+\left|x+3\right|=0\)
\(\left|x-2\right|\ge0;\left|x+3\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-2\right|=0\\\left|x+3\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-3\end{cases}vl}}\)
\(c,3x^2+9x+6=0\)
\(\Rightarrow3x^2+3x+6x+6=0\)
\(\Rightarrow3x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Rightarrow\left(3x+6\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+6=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)
\(d,x^2-7x-8=0\)
\(\Rightarrow x^2+x-8x-8=0\)
\(\Rightarrow x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Rightarrow\left(x-8\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)
*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)
\(M=x^2+11xy-y^2\)
* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)
Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)
=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Thay x = 5/2 ; y = -4/3 vào M ta được :
\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)
\(M=\frac{-1159}{36}\)
Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3
Không chắc nha
\(a,\left(a+2\right)^2+\left|3b-4\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left(a+2\right)^2=0\\\left|3b-4\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}a+2=0\\3b-4=0\end{cases}}\Rightarrow\hept{\begin{cases}a=-2\\b=\frac{4}{3}\end{cases}}\)
\(b,\left(3a^2+27\right)^2+\left|b-1\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left(3a^2+27\right)^2=0\\\left|b-1\right|=0\end{cases}\Rightarrow\hept{\begin{cases}3a^2+27=0\\b-1=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a^2=27\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a^2=9\\b=1\end{cases}\Rightarrow}\hept{\begin{cases}\orbr{\begin{cases}a=3\\a=-3\end{cases}}\\b=1\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a^2=27\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a^2=9\\b=1\end{cases}\Rightarrow}\hept{\begin{cases}\orbr{\begin{cases}a=3\\a=-3\end{cases}}\\b=1\end{cases}}}\)
\(\frac{3}{4}x-\frac{2}{3}.\left(\frac{3}{5}x-\frac{6}{5}\right)=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{3}{4}x-\frac{2}{5}x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\left(\frac{3}{4}-\frac{2}{5}\right)x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\left(\frac{15}{20}-\frac{8}{20}\right)x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{7}{20}x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{1}{7}-\frac{4}{5}=\frac{2}{9}x-\frac{7}{20}x\)
\(\frac{5}{35}-\frac{28}{35}=\left(\frac{2}{9}-\frac{7}{20}\right)x\)
\(\frac{-23}{35}=\left(\frac{40}{180}-\frac{63}{180}\right)x\)
\(\frac{-23}{180}x=\frac{-23}{35}\)
\(x=\frac{-23}{35}:\frac{-23}{180}\)
\(x=\frac{-23}{35}.\frac{180}{-23}\)
\(x=\frac{180}{35}\)
Vậy \(x=\frac{180}{35}\)
Chúc bạn học tốt
Bài 2:
Giải:
Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow x=5k,y=4k\)
Ta có: \(x^2-y^2=1\)
\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)
\(\Rightarrow5^2.k^2-4^2.k^2=1\)
\(\Rightarrow k^2\left(5^2-4^2\right)=1\)
\(\Rightarrow k^2.9=1\)
\(\Rightarrow k^2=\frac{1}{9}\)
\(\Rightarrow k=\pm\frac{1}{3}\)
+) \(k=\frac{1}{3}\Rightarrow x=\frac{5}{3};y=\frac{4}{3}\)
+) \(k=\frac{-1}{3}\Rightarrow x=\frac{-5}{3};y=\frac{-4}{3}\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(\frac{5}{3};\frac{4}{3}\right);\left(\frac{-5}{3};\frac{-4}{3}\right)\)
Bài 3:
Giải:
Ta có: \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{a}{21}=\frac{b}{14}\)
\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{21}=\frac{c}{15}\)
\(\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{15}\)
...
Bài 4:
Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)
\(\Rightarrow a=2k,b=3k,c=5k\)
Ta có: \(P=\frac{b+c-a}{a-b+c}=\frac{3k+5k-2k}{2k-3k+5k}=\frac{\left(3+5-2\right)k}{\left(2-3+5\right)k}=\frac{6}{4}=\frac{3}{2}\)
Vậy \(P=\frac{3}{2}\)
a, Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)'
Ta có: \(\frac{a}{b}=\frac{bk}{b}=k\left(1\right)\)
\(\frac{3a+2c}{3b+2d}=\frac{3bk+2dk}{3b+2d}=\frac{k\left(3b+2d\right)}{3b+2d}=k\left(2\right)\)
Từ (1) và (2) => đpcm
b, Đặt a/b=c/d=k => a=bk,c=dk
Ta có: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\left(2\right)\)
Từ (1) và (2) => đpcm
Vì \(\left(3a-4\right)^{500}\ge0\)\(\forall a\inℝ\); \(\left(9-3b\right)^{600}\ge0\)\(\forall b\inℝ\)
\(\Rightarrow\left(3a-4\right)^{500}+\left(9-3b\right)^{600}\ge0\)\(\forall a,b\inℝ\)
Mà \(\left(3a-4\right)^{500}+\left(9-3b\right)^{600}\le0\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3a-4=0\\9-3b=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=4\\3b=9\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{4}{3}\\b=3\end{cases}}\)
Vậy....
anh làm còn thiếu nhưng mà thôi .
:>