Violympic toán 7

bn có ghi thiếu đề ko?

a) \(\dfrac{1}{7}.\dfrac{-3}{8}+\dfrac{1}{7}.\dfrac{-13}{8}\)

\(=\dfrac{1}{7}.\left(\dfrac{-3}{8}+\dfrac{-13}{8}\right)\)

\(=\dfrac{1}{7}.\left(-2\right)\)

\(=\dfrac{-2}{7}\)

b) \(\dfrac{3}{5}.\dfrac{13}{46}-\dfrac{1}{10}.\dfrac{16}{23}\)

\(=\dfrac{39}{230}\) - \(\dfrac{8}{115}\)

\(=\dfrac{1}{10}\)

a, \(\dfrac{1}{7}.\left(-\dfrac{3}{8}\right)+\dfrac{1}{7}.\left(-\dfrac{13}{8}\right)\)

=> \(\dfrac{1}{7}.\left[\left(-\dfrac{3}{8}\right)+\left(-\dfrac{13}{8}\right)\right]\)

=> \(\dfrac{1}{7}.-\dfrac{16}{8}\)

=> \(\dfrac{1}{7}.\left(-2\right)\)

=> \(\dfrac{-2}{7}\)

b/ \(\dfrac{1}{7}\) . ( \(\dfrac{-3}{8}\)) +\(\dfrac{1}{7}\).(\(\dfrac{-13}{8}\))

= \(\dfrac{1}{7}\) . (\(\dfrac{-3}{8}\) + \(\dfrac{-13}{8}\) )

= \(\dfrac{1}{7}\). (-2)

=\(\dfrac{-2}{7}\)

b/ \(\dfrac{3}{5}\).\(\dfrac{13}{46}\)-\(\dfrac{1}{10}\) .\(\dfrac{16}{23}\)

= \(\dfrac{39}{230}\)-\(\dfrac{8}{115}\)

=\(\dfrac{1}{10}\)

Hàm số đồ thị y= -2 x

cho x= -1 thì y= 2

Thay x=-1 thì y=2 vào hầm số y=-2x, ta đc A(-1;2)

Đồ thị hàm số y=-2x là đường thẳng OA

Vẽ hình ko đc đẹp cho lắm nha!

\(3^{\left(n+2\right)\left(n-1\right)}=1\)

\(\Rightarrow\)\(\left(n+2\right)\left(n-1\right)=0\)

\(\Rightarrow n+2=0\Rightarrow n=-2\) hoặc\(n-1=0\Rightarrow n=1\)

thiếu đề chắc rồi,

đề tau ghi thiếu

Thảo nào làm ko ra

a, Theo đề ta có:

\(2.3^x-405=3^{x-1}\)

=> \(2.3^x-405=3^x:3\)

=> \(405=(2.3^x)-(3^x:3)\)

=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)

=> \(405=3^x(2-\dfrac{1}{3})\)

=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)

=> \(405=3^x.\dfrac{5}{3}\)

=> \(3^x=405:\dfrac{5}{3}\)

=>\(3^x=405.\dfrac{3}{5}\)

=> \(3^x=81.3\)

=> \(3^x=243\)

=> \(3^x=3^5\)

=> x=5

Vậy:..............................

a,

\(\dfrac{1}{4}x-1+\dfrac{1}{3}\left(\dfrac{5}{2}x-7\right)-\left(\dfrac{5}{8}x-2\right)=\dfrac{7}{2}\)

\(\Rightarrow\dfrac{1}{4}x-1+\dfrac{5}{6}x-\dfrac{7}{3}-\dfrac{5}{8}x+2=\dfrac{7}{2}\)

\(\Rightarrow\dfrac{1}{4}x+\dfrac{5}{6}x-\dfrac{5}{8}x=\dfrac{7}{2}+1+\dfrac{7}{3}-2\)

\(\Rightarrow\dfrac{11}{24}x=\dfrac{29}{6}\)

\(\Rightarrow x=\dfrac{116}{11}\)

b,

\(\left|2-\dfrac{3}{2}x\right|-4=x+2\)

\(\Rightarrow\left|2-\dfrac{3}{2}x\right|=x-2\)

\(\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-\left(x+2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-\dfrac{3}{2}x=x+2\\2-\dfrac{3}{2}x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2-2=x+\dfrac{3}{2}x\\2+2=-x+\dfrac{3}{2}x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=0\\\dfrac{1}{2}x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

c,

\(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)

\(\Rightarrow-\dfrac{6}{5}x+\dfrac{3}{5}-x^2+\dfrac{1}{2}x=\dfrac{1}{6}-x^2\)

\(\Rightarrow-\dfrac{7}{10}x=\dfrac{1}{6}-\dfrac{3}{5}-x^2+x^2\)

\(\Rightarrow-\dfrac{7}{10}x=-\dfrac{13}{30}\Leftrightarrow x=\dfrac{13}{21}\)

mình bt nè

\(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\)

\(\Rightarrow\dfrac{2\left(x+y\right)}{30}=\dfrac{5\left(y+z\right)}{30}=\dfrac{3\left(z+x\right)}{30}\)

\(\Rightarrow\dfrac{x+y}{15}=\dfrac{y+z}{6}=\dfrac{z+x}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+y}{15}=\dfrac{y+z}{6}=\dfrac{z+x}{10}=\dfrac{z+x-y-z}{10-6}=\dfrac{x-y}{4}\left(1\right)\)

\(\dfrac{x+y}{15}=\dfrac{y+z}{6}=\dfrac{z+x}{10}=\dfrac{x+y-z-x}{15-10}=\dfrac{y-z}{5}\left(2\right)\)

Từ (1) và (2) ta có đpcm

Nghĩ là A hoặc C

Hình B vi :

Tính theo hàng dọc, mỗi hàng đều có 2 hình giống nhau nên ? sẽ là hình B

mình thấy đáp án B và D nó hơi giống nhau một tí ???