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b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
\(2x^3-1=15\)
\(\Rightarrow2x^3=16\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
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 \(B=100\)
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
\(a,\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)và x + y + z = 49
Ta có : \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{2}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{2}+\frac{5}{4}}=\frac{49}{\frac{19}{4}}=49\cdot\frac{4}{19}=\frac{196}{19}\)
Vậy : \(\hept{\begin{cases}\frac{x}{\frac{3}{2}}=\frac{196}{19}\\\frac{y}{\frac{4}{2}}=\frac{196}{19}\\\frac{z}{\frac{5}{4}}=\frac{169}{14}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{294}{19}\\y=\frac{392}{19}\\z=\frac{245}{19}\end{cases}}\)
\(b,\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\)và 2x + 3y - z = 186
Ta có : \(\frac{x}{y}=\frac{3}{4};\frac{y}{z}=\frac{5}{7}\Leftrightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
Vậy : \(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}\)
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 \(z=41;x=2;y=57\)
\(4x^3-3=29\Rightarrow x^3=\dfrac{29+3}{4}=8\Rightarrow x=\sqrt[3]{8}=2\)
Thay số: \(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\)
Suy ra: \(y=\left(-16\right)\cdot2+25\Leftrightarrow y=-7\) và \(z=25\cdot2-49\Leftrightarrow z=1\)
\(A=x+2y+3z\Leftrightarrow2+\left(-14\right)+3=-9\)
\(4x^3-3=29\Rightarrow x^3=\dfrac{32}{4}=2^3\Rightarrow x=3\)
\(\dfrac{19}{9}=\dfrac{2y-2.25}{-32}=\dfrac{3z+49.3}{75}=\dfrac{2y+3z+49.3-25.2}{75-32}=\dfrac{2y+3z+97}{43}\)
\(\dfrac{\left(2y+3z+3\right)+94}{43}=\dfrac{19}{9}\) \(\Rightarrow\left(x+2y+3z\right)=\dfrac{43.19}{9}-94\)
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.