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Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}=\dfrac{\left(y^2-x^2\right)-\left(y^2+x^2\right)}{3+5}=\dfrac{\left(y^2-x^2\right)-\left(y^2-x^2\right)}{3-5}\Rightarrow\dfrac{2y^2}{8}=\dfrac{-2x^2}{-2}\Rightarrow\dfrac{y^2}{4}=x^2\Rightarrow y^2=4x^2\)
Ta có: \(x^{10}.y^{10}=x^{10}.\left(4x^2\right)^5=1024.x^{20}=1024\Rightarrow x^{20}=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\Rightarrow y^2=4\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\) và \(y\in\left\{4;-4\right\}\)
\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}\)
\(\Leftrightarrow5\left(y^2-x^2\right)=3\left(y^2+x^2\right)\)
\(\Leftrightarrow5y^2-5x^2=3y^2+3x^2\)
\(\Leftrightarrow2y^2=8x^2\)
\(\Leftrightarrow y^2=4x^2\)
\(\Leftrightarrow y^{10}=1024.x^{10}\)
Mà \(x^{10}.y^{10}=1024\)
\(\Leftrightarrow x^{10}.1024x^{10}=1024\)
\(\Leftrightarrow x^{20}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
+)Với \(x=1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
+) Với \(x=-1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy...
\(\dfrac{x-y}{x+y}=\dfrac{3}{7}\)
\(\Leftrightarrow7x-7y=3x+3y\)
=>4x=10y
=>2x=5y
hay x/5=y/2
Đặt x/5=y/2=k
=>x=5k; y=2k
\(x^2y^2=1600\)
\(\Leftrightarrow10k^2=1600\)
\(\Leftrightarrow k^2=160\)
TH1: \(k=4\sqrt{10}\)
\(x=20\sqrt{10};y=8\sqrt{10}\)
TH2: \(k=-4\sqrt{10}\)
\(x=-20\sqrt{10};y=-8\sqrt{10}\)
Lời giải:
\(\frac{x-y}{x+y}=\frac{3}{7}\Rightarrow 7(x-y)=3(x+y)\)
\(\Leftrightarrow 4x=10y\Rightarrow y=0,4x\)
Lại có: \(x^3y^3=1000\Leftrightarrow (xy)^3=1000\Rightarrow xy=\sqrt[3]{1000}=10\)
Thay \(y=0,4x\) ta có:
\(x.0,4x=10\Leftrightarrow x^2=25\Rightarrow x=\pm 5\)
Nếu \(x=5\rightarrow y=0,4x=2\)
Nếu \(x=-5\rightarrow y=0,4x=-2\)
ta có x^3.y^3=(x.y)^3=1000
<=>(x.y)^3=10^3
<=>x.y=10
ta có (x-y)/(x+y)=3/7 <=> 7x-7y=3x+3y
<=> 4x=10y
<=>x=y.5/2
thay x= y.5/2 vào x.y=10 ta có:
y.5/2.y=10
<=>y^2=4
<=>y=2 hoặc y=-2
với y=2 ta có x=5, với y=-2 ta có x=-5
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
2. Tính:
a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)
=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)
=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)
=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)
b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)
=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)
=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)
=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)
=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)
=\(\dfrac{13}{4}\)
3. Tìm x
a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)
\(\left(x-5\right).\left(x-5\right)=8.18\)
\(\left(x-5\right)^2=144\)
\(x-5=\sqrt{144}\)
\(x-5=12\)
\(x=12+5\)
\(x=17\)
b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)
\(x^{10}-2^{10}=x^8-2^8\)
\(x^{10}+x^8=2^{10}+2^8\)
\(\Rightarrow x=2\)
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
1: Vì x^2 >=0 với mọi x ; (y- 1/10)^4 >=0 với mọi y
==> x^2 + (y- 1/10)^4 >= 0.
Do đó dấu = xảy ra tức là x^2 + (y- 1/10)^4 =0 <=> x^2 =0 và (y- 1/10)^4 =0 <=> x=0; y=1/10
bài 2 kiểu tương tự nha
(x - 1 )^4sẽ \(0\le\left(x-1\right)^4\)
(y+2)^100 sẽ \(0\le\left(y+2\right)^{100}\)
đến đó bn làm nhé
a) Vì \(x^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow x^2=0;y-\dfrac{1}{10}=0\)
\(\Rightarrow x=0;y=\dfrac{1}{10}\)
b) Vì \(\left(x-1\right)^4\ge0\forall x\)
\(\left(y+2\right)^{100}>0\forall y\)
\(\Rightarrow\left(x-1\right)^4+\left(y+2\right)^{100}\ge0\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow x-1=0;y+2=0\)
\(\Rightarrow x=1;y=-2\)
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
Lời giải:
Ta có: \(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}\Rightarrow 5(y^2-x^2)=3(y^2+x^2)\)
\(\Rightarrow 2y^2=8x^2\Rightarrow y^2=4x^2\)
\(\Rightarrow y^{10}=4^5x^{10}=(2x)^{10}\)
Do đó:
\(x^{10}y^{10}=x^{10}.(2x)^{10}=1024\)
\(\Leftrightarrow (2x^2)^{10}=1024=2^{10}=(-2)^{10}\)
\(\Rightarrow \left[\begin{matrix} 2x^2=2\\ 2x^2=-2(\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow x^2=1\Rightarrow x=\pm 1\)
\(y^2=4x^2=4\Rightarrow y=\pm 2\)
Vậy \((x,y)=(1,-2); (1,2); (-1,2); (-1,-2)\)