99 x 8,27 + 8,27

=99 x 8,27 + 8,27 x 1

=(99+1) x 8,27

=100 x 8,27

=827

99 x 8,27 + 8,27 

= ( 99 x 8,27 ) + ( 8,27 x 1 )

= 8,27 x ( 99 + 1 )

= 8,27 x 100 

= 827 

@Trunglaai?

99 x 8,27+8,27=99x8,27+8,27x1=8,27x(99+1)=8,27 x 100=827 nha bạn

=827

Tính bằng cách thuận tiện nhất 

99 x 8,27 +8,27

= 8,27 x (99+1)

=  8,27  x 100

= 827 

=99 x 8,27 +8,27 x 1

=8,27 x (99+1)

=8,27 x 100

=827

=8,27+7,16+9,33-7,27-6,16-8,33

=(8,27-7,27)+(7,16-6,16)+(9,33-8,33)

=        1        +       1        +  1

=                           3

3

a) X : \(\dfrac{2}{7}\) = 8,27 + 19,73

X : \(\dfrac{2}{7}\) = 28

X = 28 x \(\dfrac{2}{7}\)

X = 8

b) 8,15 x X + 1,85 x X = 124

X x ( 8,15 + 1,85 ) = 124

X x 10 = 124

X = 124 : 10

X = 12,4

câu a mình chia dư.

cũng được bạn ạ

a, x - 8, 27=1,645

x=1,645+8,27

x=9,915

b,4,8:x=9,6

x=4,8:9,6

x=0,5

nhớ tk đúng cho mình nhé ,ko bố mai đến lớp bố đập chết mày

Câu 1.

Khi mở khóa K:

\(I_m=I_1=0,4A\)

Khi đóng khóa K:

\(I_m=I_1+I_2=0,6\Rightarrow I_2=0,2A\)

\(U_1=0,4\cdot5=2V\)

\(\Rightarrow U_2=U_1=2V\)

\(\Rightarrow U=U_1=U_2=2V\)

\(R_2=\dfrac{U_2}{I_2}=\dfrac{2}{0,2}=10\Omega\)

-Bài 3:

2) -Áp dụng BĐT Caushy Schwarz ta có:

\(A=\dfrac{1}{x^3+3xy^2}+\dfrac{1}{y^3+3x^2y}\ge\dfrac{\left(1+1\right)^2}{x^3+3xy^2+3x^2y+y^3}=\dfrac{4}{\left(x+y\right)^3}\ge\dfrac{4}{1^3}=4\)-Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

Câu 59:B

Câu 63: A