Ta có: \(\frac{5}{x}-\frac{y}{4}=\frac{1}{8}\)

=> \(\frac{5}{x}=\frac{1}{8}+\frac{y}{4}\)

=> \(\frac{5}{x}=\frac{1+2y}{8}\)

=> (1 + 2y)x = 40 = 1 . 40 = 2.20 = 5 . 8 = 4 . 10

Vì 1 + 2y là số lẽ nên => 1 + 2y \(\in\)1; 5;-1;-5

Lập bảng : 

Vậy ...

b) Ta có: \(\frac{x}{5}+\frac{1}{10}=\frac{1}{y}\)

=> \(\frac{2x+1}{10}=\frac{1}{y}\)

=> (2x + 1).y = 10 = 1 . 10 = 2. 5

Vì 2x + 1 là số lẽ => 2x + 1 \(\in\){1; 5; -1; -5}

Lập bảng: tương tự câu a

c) Như câu b.

\(\text{Ta có: }\) \(\frac{1}{4}x+\frac{1}{8}x+\frac{1}{16}x=1\)

\(\Rightarrow\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)x=1\)

\(\Rightarrow\frac{7}{16}x=1\)

\(\Rightarrow x=1:\frac{7}{16}\)

\(\Rightarrow x=\frac{16}{7}\)

Tìm x :

a, 1/4 * x + 1/8 * x + 1/16 * x = 1

x * ( 1/4 + 1/8 + 1/16 ) = 1

x * 7/16 = 1

     x = 1 : 7/16

     x = 16/7

b, 1/5 + 1/3 x ( x + 1 ) = 1/4

            1/3 x ( x + 1 ) = 1/4 - 1/5

            1/3 x ( x + 1 ) = 1/20

                      x + 1   = 1/20 : 1/3

                      x + 1 = 3/20

                      x      = 1 - 3/20 

                     x      = 7/20

Tính nhanh :

1/5 x 27 + 1/5 x 33 + 1/5 x 40

= 1/5 x ( 27 + 33 + 40 )

= 1/5 x 100

= 20 

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

|x-1| + |4-x| = 3

Áp dụng bất đẳng thức ta có:

 |x-1| + |4-x |  \(\ge\)|x-1+ 4-x|  = 3

Dấu = xảy ra khi và chỉ khi :  (x-1)(4-x) \(\ge\)0

\(\Rightarrow\) 1\(\le\)x \(\le\)4

Vậy 1\(\le\)x \(\le\)4 là giá trị cần tìm

có x ko thì lm sao tìm y đc??

Chỉ tìm x thôi bạn nhé......mk ghi lộn :P Mong bạn giúp

x=17 nhé

mk kô bt đánh ps đâuu à