Tìm Y biết: 2+3/4-1/2-y = 7
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a) y × 2/5 + y × 3/5 = 5/7
y x ( 2/5 + 3/5) = 5/7
y x 5/5 = 5/7
y x 1 = 5/7
y = 5/7 : 1
y = 5/7
b) y × 7/2 - y × 3/2 = 3/4
y x ( 7/2 - 3/2) = 3/4
y x 4/2 = 3/4
y x 2 = 3/4
y = 3/4 : 2
y = 3/4 x 1/2
y = 3/8
`a,yxx(2/5+3/5)=5/7`
`y xx1=5/7`
`y=5/7:1`
`y=5/7`
`b,y xx(7/2-3/2)=3/4`
`y xx2=3/4`
`y =3/4:2`
`y=3/2`
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
Đề này cũng sai rồi, cho mình xin phép sửa:
Tìm x,y,z biết: \(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\) và \(x+2=y+3=z+4\)
Bài làm:
Ta có: \(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\)
=> \(x=\frac{1}{2}\) hoặc \(y=-\frac{1}{3}\) hoặc \(z=2\)
+ Nếu: \(x=\frac{1}{2}\Rightarrow\hept{\begin{cases}y=\frac{1}{2}+2-3=-\frac{1}{2}\\z=\frac{1}{2}+2-4=-\frac{3}{2}\end{cases}}\)
+ Nếu: \(y=-\frac{1}{3}\Rightarrow\hept{\begin{cases}x=-\frac{1}{3}+3-2=\frac{2}{3}\\z=-\frac{1}{3}+3-4=-\frac{4}{3}\end{cases}}\)
+ Nếu: \(z=2\Rightarrow\hept{\begin{cases}x=2+4-2=4\\y=2+4-3=3\end{cases}}\)
Vậy ta có 3 cặp số (x;y;z) thỏa mãn: \(\left(\frac{1}{2};-\frac{1}{2};-\frac{3}{2}\right);\left(\frac{2}{3};-\frac{1}{3};-\frac{4}{3}\right);\left(4;3;2\right)\)
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.
\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)
Vậy \(\left(x;y\right)=\left(5;-2\right)\)
Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
2 + 3/4 - 1/2 - y = 7
9/4 - y = 7
y = 9/4 - 7
y = -19/4
\(2+\frac{3}{4}-\frac{1}{2}-y=7\)
\(\frac{8}{4}+\frac{3}{4}-\frac{2}{4}-y=7\)
\(\frac{9}{4}-y=7\)
\(y=\frac{9}{4}-7=\frac{9}{4}-\frac{28}{4}\)
\(\Rightarrow x=-\frac{19}{4}\)