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Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
Bài 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
\(\frac{12}{-6}=\frac{x}{5}=\frac{-y}{3}=\frac{z}{-17}=\frac{-t}{-9}\)
\(-6x=12\cdot5=60\Rightarrow x=-10\)
\(-y\cdot\left(-6\right)=12\cdot3=36\Rightarrow y=6\)
\(-6z=-17\cdot12=>z=34\)
\(-t\cdot\left(-6\right)=-9\cdot12=>t=-18\)
a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)
\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)
\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)
câu cuối làm tương tự
Do 12/-6=x/5 nên 12.5/-6.x=>x =12.5/-6 => x=-10. Vậy x = -10
Do 12/-6=-y/3 nên 12.3=-6.-y=>-y=12.3/-6=>-y=-6=>y=6. Vậy y=6
Do 12/-6=z/-17 nên 12.-17=-6.z=>z=12.-17/-6=>z=34. Vậy z=34
Do 12/-6=-t/-9 nên 12. -9=-6. t=>t=12.-9/-6+>t=18. Vậy t=18
CHÚC BẠN HỌC GIỎI
TK MÌNH NHÉ
\(\frac{-24}{-6}=4\)
\(\Rightarrow\frac{x}{5}=4\Rightarrow x=5\times4=20\)
\(\frac{-y}{3}=4\Rightarrow y=-3\times4=-12\)
\(\frac{-t}{-9}=4\Rightarrow t=4\times9=36\)
Vậy x=20; y=-12; z=36
mình mới học lớp 5
duyệt n ha