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Do \(\left(x-\frac{1}{3}\right)^{46}\ge0,\left(3-4y\right)^{100}\ge0,\left|z-5\right|\ge0\)nên \(\left(x-\frac{1}{3}\right)^{46}+\left(3-4y\right)^{100}+\left|z-5\right|\ge0\)
Để \(\left(x-\frac{1}{3}\right)^{46}+\left(3-4y\right)^{100}+\left|z-5\right|=0\)thì \(\hept{\begin{cases}x-\frac{1}{3}=0\\3-4y=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{3}{4}\\z=5\end{cases}}}\)
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\(A=\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)\left(\frac{1}{36}-1\right)...\left(\frac{1}{100}-1\right)\)
\(-A=\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{100}\right)\)
\(-A=\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\cdot...\cdot\frac{99}{100}\)
\(-A=\frac{\left(3\cdot5\right)\left(4\cdot6\right)\left(5\cdot7\right)...\left(9\cdot11\right)}{\left(4\cdot4\right)\left(5\cdot5\right)\left(6\cdot6\right)...\left(10\cdot10\right)}\)
\(-A=\frac{\left(3\cdot4\cdot5\cdot...\cdot9\right)\left(5\cdot6\cdot7\cdot...\cdot11\right)}{\left(4\cdot5\cdot6\cdot...\cdot10\right)\left(4\cdot5\cdot6\cdot...\cdot10\right)}\)
\(-A=\frac{3\cdot11}{10\cdot4}=\frac{33}{40}\)
\(A=-\frac{33}{40}\)
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Ta có:\(\frac{x}{4}=\frac{y}{7}\Rightarrow\frac{x^2}{16}=\frac{y^2}{49}\)
\(=\frac{3x^2}{48}=\frac{4y^2}{196}=100\)
\(\Rightarrow\hept{\begin{cases}x^2=1600\\y^2=4900\end{cases}}\)
TH1:x=40,y=70
TH2:x=40,y=-70
TH3:x=-40,y=70
TH4:x=-40,y=-70
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a: 2x-3y-4z=24
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)
=>x=-6/7; y=-36/7; z=-18/7
b: 6x=10y=15z
=>x/10=y/6=z/4=k
=>x=10k; y=6k; z=4k
x+y-z=90
=>10k+6k-4k=90
=>12k=90
=>k=7,5
=>x=75; y=45; z=30
d: x/4=y/3
=>x/20=y/15
y/5=z/3
=>y/15=z/9
=>x/20=y/15=z/9
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)
=>x=500; y=375; z=225
Thiếu đề à cậu????