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a, Ta có: \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\Leftrightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}\) và \(5x+y-2z=28\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\)
+) \(\dfrac{5x}{50}=2\Rightarrow5x=100\Rightarrow x=20\)
+) \(\dfrac{y}{6}=2\Rightarrow y=12\)
+) \(\dfrac{2z}{42}=2\Rightarrow2z=84\Rightarrow z=42\)
Vậy ...
b, Ta có:
\(3x=2y\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
\(7y=5z\Leftrightarrow\dfrac{y}{5}=\dfrac{z}{7}\)
Ta lại có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\) và \(x-y+z=32\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)
+) \(\dfrac{x}{10}=2\Rightarrow x=20\)
+) \(\dfrac{y}{15}=2\Rightarrow y=30\)
+) \(\dfrac{z}{21}=2\Rightarrow z=42\)
Vậy ...
2) Ta có: \(\frac{x_1}{y_2}=\frac{x_2}{y_1}\Rightarrow\frac{x_1^2}{y_2^2}=\frac{x_2^2}{y_1^2}=\frac{x_1^2+x_2^2}{y_1^2+y_2^2}=\frac{2^2+3^2}{52}=\frac{1}{4}\)
\(\Rightarrow\frac{x_1^2}{y_2^2}=\frac{1}{4}\Rightarrow y_2^2=16\Rightarrow\)\(\orbr{\begin{cases}y_2=-4\\y_2=4\end{cases}\Rightarrow}\)\(\orbr{\begin{cases}y_1=-6\\y_1=6\end{cases}}\)
=> KL....
I2x+3I=x+2
TH1: Nếu \(x\le-\frac{3}{2}\)(*), =>I2x+3I=-2x-3
PT: -2x-3=x+2 <=> x=\(-\frac{5}{3}\)(tm (*))
TH2: Nếu \(x>-\frac{3}{2}\)(**), => I2x+3I=2x+3
PT: 2x+3=x+2 => x=-1 (tm (**))
Vậy x=...
\(1)-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)
\(\Rightarrow-4x^2-\left(-20x\right)-16x+4x^2=-3\)
\(\Rightarrow20x-14x=-3\)
\(\Rightarrow6x=-3\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(2)\) Theo bài ra, ta có: \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\)
\(\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)
\(\Rightarrow\sqrt[3]{\left(\dfrac{x}{2}\right)^3}=\sqrt[3]{\left(\dfrac{y}{4}\right)^3}=\sqrt[3]{\left(\dfrac{z}{6}\right)^3}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{6}\right)^2\)
\(\Rightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Suy ra:
\(+)\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}.4=1=\left(\pm1\right)^2\Rightarrow x=\pm1\)
\(+)\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{16}.4=\dfrac{1}{4}=\left(\pm\dfrac{1}{2}\right)^2\Rightarrow y=\pm\dfrac{1}{2}\)
\(+)\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{36}.4=\dfrac{1}{9}=\left(\pm\dfrac{1}{3}\right)^2\Rightarrow z=\pm\dfrac{1}{3}\)
Vậy \(\left(x;y;z\right)\in\left\{\left(-1;-\dfrac{1}{2};-\dfrac{1}{3}\right);\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\right\}\)
Bài 2 )
\(a\left(y+z\right)=b\left(x+z\right)=c\left(x+y\right)\)
\(\Leftrightarrow\frac{a\left(y+z\right)}{abc}=\frac{b\left(x+z\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\frac{y+z}{bc}=\frac{x+z}{ac}=\frac{x+y}{ab}\)
\(\Leftrightarrow\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}\)
Đặt \(\frac{bc}{y+z}=\frac{ac}{x+z}=\frac{ab}{x+y}=k\)
\(\Rightarrow\left\{\begin{matrix}bc=k\left(y+z\right)=ky+kz\\ac=k\left(x+z\right)=kx+kz\\ab=k\left(x+y\right)=kx+ky\end{matrix}\right.\) (1)
Gỉa sử điều cần chứng minh là đúng ta có
\(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\frac{y-z}{ab-ac}=\frac{z-x}{bc-ab}=\frac{x-y}{ac-bc}\)
Thế (1) vào biểu thức
\(\frac{y-z}{kx+ky-\left(kx+kz\right)}=\frac{z-x}{ky+kz-\left(kx+ky\right)}=\frac{x-y}{kx+kz-\left(ky+kz\right)}\)
\(\Leftrightarrow\frac{y-z}{ky-kz}=\frac{z-x}{kz-kx}=\frac{x-y}{kx-ky}\)
\(\Leftrightarrow\frac{y-z}{k\left(y-z\right)}=\frac{z-x}{k\left(z-x\right)}=\frac{x-y}{k\left(x-y\right)}\)
\(\Leftrightarrow\frac{1}{k}=\frac{1}{k}=\frac{1}{k}\) ( điều này luôn luôn đúng )
\(\Rightarrow\) ĐPCM
Bài 1: a/b=b/c=c/a chứ không phải c/d
áp dụng tính chất dãy tỉ số bằng nhau, ta có:
a/b=b/c=c/a=(a+b+c)/(b+c+a)=1
a/b=1 => a=b
b/c=1 => b=c
Vậy a=b=c