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Tính:
\(3.\left(x-2\right)-4.\left(2x+1\right)-5.\left(2x+3\right)=50\)
\(\Rightarrow3x-6-\left(8x+4\right)-\left(10x+15\right)=50\)
\(\Rightarrow3x-6-8x-4-10x-15=50\)
\(\Rightarrow-15x-25=50\)
\(\Rightarrow-15x=50+25\)
\(\Rightarrow-15x=75\)
\(\Rightarrow x=75:\left(-15\right)\)
\(\Rightarrow x=-5.\)
Vậy \(x=-5.\)
Chúc bạn học tốt!
\(3x^2-3xy-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+x-y-6x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+\left(x-y\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x-2=-22\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-\left(6x+2\right)=-22\)
\(\Rightarrow\left(x-y\right)\left(3x+1\right)-2\left(3x+1\right)=-22\)
\(\Rightarrow\left(3x+1\right)\left(x-y-2\right)=-22\)
Ta có bảng sau:
| \(3x+1\) | \(-1\) | \(1\) | \(-22\) | \(22\) |
| \(x\) | \(x\notin Z\) | \(0\) | \(x\notin Z\) | \(7\) |
| \(x-y-2\) | \(-22\) | \(-1\) | ||
| \(y\) | \(-20\) | \(6\) |
Vậy ta có 2 bộ (x,y) là (0;-20) và (7;6)
Chúc bạn học tốt!
Mình làm một câu ví dụ thui nha
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
\(\frac{5x}{50}=2\Rightarrow x=20\)
\(\frac{y}{6}=2\Rightarrow y=12\)
\(\frac{2z}{42}=2\Rightarrow x=42\)
mấy câu khác thì tương tự
tíc mình nha bạn
\(1)-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)
\(\Rightarrow-4x^2-\left(-20x\right)-16x+4x^2=-3\)
\(\Rightarrow20x-14x=-3\)
\(\Rightarrow6x=-3\)
\(\Rightarrow x=-\dfrac{1}{2}\)
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 \(\left(x;y;z\right)\in\left\{\left(-1;-\dfrac{1}{2};-\dfrac{1}{3}\right);\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\right\}\)