Tìm x ∈ Z, biết: x + (-23) = (-100) + 77.
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\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)
\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)
\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)
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Nếu x \(\ge\) 0
x + 1 + x + 2+ ....... +x + 100 = 101x
100x + 100 x 101 : 2 = 101x
x = 100 x 101 : 2
x = 5050
Nếu x < 0
-(x + 1) - (x + 2) -.... - (x + 100) = 101x
-x - 1 - x - 2 - .... - x - 100 = 101x
- 100x - (100 x 101 : 2) = 101x
-100x - 5050= 101x
101x + 100x = -5050
201x = -5050
x = -5050/201 (loại , vì không phải số nguyên)
Vậy x = 5050
Nếu x $\ge$≥ 0
x + 1 + x + 2+ ....... +x + 100 = 101x
100x + 100 x 101 : 2 = 101x
x = 100 x 101 : 2
x = 5050
Nếu x < 0
-(x + 1) - (x + 2) -.... - (x + 100) = 101x
-x - 1 - x - 2 - .... - x - 100 = 101x
- 100x - (100 x 101 : 2) = 101x
-100x - 5050= 101x
101x + 100x = -5050
201x = -5050
x = -5050/201 (loại , vì không phải số nguyên)
Vậy x = 5050
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Theo tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{18x-27y}{100}=\frac{27y-24z}{101}=\frac{24z-18x}{102}=\frac{18x-27y+27y-24z+24z-18x}{100+101+102}=\frac{0}{303}=0\)
\(\Rightarrow\frac{27y-24z}{101}=0\Rightarrow27y-24z=0\Rightarrow27y=24z\Rightarrow9y=8z\Rightarrow\frac{y}{8}=\frac{z}{9}\) (1)
\(\frac{24z-18x}{102}=0\Rightarrow24z-18x=0\Rightarrow18x=24z\Rightarrow3x=4z\Rightarrow\frac{x}{4}=\frac{z}{3}\Rightarrow\frac{x}{12}=\frac{z}{9}\) (2)
Từ (1) và (2) suy ra \(\frac{x}{12}=\frac{y}{8}=\frac{z}{9}=\frac{x+y+z}{12+8+9}=\frac{116}{29}=4\)
=> x/12 = 4 => x = 48
y/8 = 4 => y = 32
z/9 = 4 => z = 36
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x+(-23)=(-100)+77
x+(-23)=-(100-77)
x+(-23)=-23
x =-23-(-23)
x =0
x + ( -23) = (-100) + 77
<=> x = (-100) + 77 + 23
<=> x = 0 (t/m)
Ta có: x + (-23) = (-100) + 77
x + (-23) = - (100 – 77)
x + (-23) = -23
x + (-23) = 0 +(-23)
Nên x = 0