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\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{y-z-25-9}{16-25} \)
\(<=>\frac{x+16}{9}=\frac{2x^3-34}{-9} \)
<=>\(-x-16=2x^3-34\)
<=>\(2x^3+x-18=0\)
=> x=2
=>\(\frac{2+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=2\)
=>y=57
=>z=41
Ta có:\(2x^3-1=15\Rightarrow x^3=8\Rightarrow x=2\)
\(\frac{y-25}{16}=2\Rightarrow y=2.16+25=57\)
\(\frac{z+9}{25}=2\Rightarrow z=25.2-9=41\)
\(2x^3-1=15\)
\(2x^3=16\)
\(x^3=8\)
\(x=2\)
\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\)
\(\Rightarrow\frac{y-25}{16}=2\)
\(\Rightarrow y-25=32\)
\(\Rightarrow y=57\)
\(\Leftrightarrow\frac{z+9}{25}=2\)
\(\Rightarrow z+9=50\)
\(\Rightarrow z=50-9=41\)
Vậy \(z=41;x=2;y=57\)
\(2x^3-1=15\)
\(\Rightarrow2x^3=16\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
Thay x vào \(\dfrac{x+16}{9}=\dfrac{y-25}{16}+\dfrac{z+9}{25}\) thì tìm được y và z
Tính nốt x + y + z
\(2x^3-1=15\)
\(2x^3=16\)
\(x^3=8\)
\(\Rightarrow x=2\)
\(\dfrac{x+16}{9}=\dfrac{y+25}{16}=\dfrac{z+9}{25}\)
\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
\(\Rightarrow\dfrac{y-25}{16}=2\)
\(\Rightarrow y-25=32\)
\(\Rightarrow y=57\)
\(\Rightarrow\dfrac{z+9}{25}=2\)
\(\Rightarrow z+9=50\)
\(\Rightarrow z=41\)
\(\Rightarrow\)\(x=2\) , \(y=57\) , \(z=41.\)
\(B=x+y+z\)
\(B=2+57+41\)
\(B=100\)
Vậy \(B=100\)
Cho \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}và\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\).Tìm x+y+z
theo bài ra ta có:
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{x+16+y-25+z+9}{9+16+25}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{x+16}{9}=\dfrac{x+y+z}{50}\left(1\right)\)ta lại có:
\(\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\\ \Rightarrow\dfrac{7+2-x}{7}+\dfrac{9+2-x}{9}=2\\ \Rightarrow\left(1+\dfrac{2-x}{7}\right)+\left(1+\dfrac{2-x}{9}\right)=2\\ \Rightarrow\left(1+1\right)+\left(\dfrac{2-x}{7}+\dfrac{2-x}{9}\right)=2\\ \Rightarrow2+\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=2\\ \Rightarrow\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=0\\ \Rightarrow2-x=0\\ \Rightarrow x=2\)
thay x = 2 vào 1 ta có:
\(\Rightarrow\dfrac{2+16}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{18}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow2=\dfrac{x+y+z}{50}\\ \Rightarrow x+y+z=2.50\\ \Rightarrow x+y+z=100\)
vậy x + y + z = 100
Ta có :
\(2x^3-1=15\)
\(\Leftrightarrow2x^3=16\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x=2\)
Thay \(x=2\) zô : \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
+) \(\dfrac{y-25}{16}=2\)
\(\Leftrightarrow y-25=32\)
\(\Leftrightarrow y=57\)
+) \(\dfrac{z+9}{25}=2\)
\(\Leftrightarrow z+9=50\)
\(\Leftrightarrow z=41\)
Ta có :
\(\left\{{}\begin{matrix}x=2\\y=57\\z=41\end{matrix}\right.\) \(\Leftrightarrow x+y+z=2+57+41=100\)
Câu 2:
a) \(\sqrt{x}=5\)
\(\Leftrightarrow x=25\)
b) \(2\sqrt{x}=\sqrt{12}\)
\(\Leftrightarrow2\sqrt{x}=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3}\)
\(\Leftrightarrow x=3\)
c) \(x^2=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)
d) \(-3\sqrt{x}=-\sqrt{18}\)
\(\Leftrightarrow-3\sqrt{x}=3\sqrt{2}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{2}\)
\(\Leftrightarrow x=2\)
e) \(x^2-1=7\)
\(\Leftrightarrow x^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\)
f) \(3\sqrt{x^2}=\sqrt{9}\)
\(\Leftrightarrow3\cdot\left|x\right|=3\)
\(\Leftrightarrow\left|x\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a, Ta có: \(\left(xyz\right)^2=\dfrac{2}{7}.\dfrac{3}{2}.\dfrac{3}{7}\)\(=\dfrac{9}{49}\)
\(\Rightarrow xyz=\sqrt{\dfrac{9}{49}}=\dfrac{3}{7}.\)
\(\Rightarrow z=\dfrac{xyz}{xy}=\dfrac{3}{7}:\dfrac{2}{7}=1,5.\)
\(\Rightarrow y=1;x=\dfrac{2}{7}\).
b, Tương tự.