`1km = 1000m`

Đội công nhân đó đã sửa được số `m` :

`1000 xx 3/5 = 600 (m)`

Đội công nhân đó còn phải sửa số `m` nữa :

`1000 - 600 = 400(m)`

Đ/s..

Đổi: 1 km = 1000 m

Số mét đường sau 2 ngày đội đó sửa được là:

1000 \(\times\) `3/5` = 600 (m đường)

Số mét đường đội công nhân đó phải sửa tiếp là:

1000 - 600 = 400 (m đường)

Đáp số: 400 m đường.

Thể tích bể :

`80 xx 50 xx 50 = 200000 (cm^3) = 200(dm^3)`

Cần đổ số lít nước :

`200 xx 80%= 160 (lít)`

Vậy....

Thể tích bể cá là:

 80x50x50 = 200000 ( cm3) => 200 dm2

Cần đổ số lít nước là:

 200x80:100=160 ( lít)

                        đáp số:/............

thứ 6

6

chiều cao 4 phần 5 của bể nha các bạn

đổi 48l = 0,048 (m³)

chiều cao của bể là

0,048 : (0,5 x 0,3) = 0,32 (m)

đs....

The common method for the equation \(\sqrt{A}+\sqrt{B}=k\left(A,B,k\ge0\right)\) (k is a constant number) usually is raise each side of the equation to the power of 2:

\(\sqrt{A}+\sqrt{B}=k\) \(\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2=k^2\) \(\Leftrightarrow A+B+2\sqrt{AB}=k^2\)\(\Leftrightarrow2\sqrt{AB}=k^2-A-B\)

And you raise each side of the equation to the power of 2 again: \(2\sqrt{AB}=k^2-A-B\Leftrightarrow\left(2\sqrt{AB}\right)^2=\left(k^2-A-B\right)^2\) \(\Leftrightarrow4AB=\left(k^2-A-B\right)^2\) 

And now we have eliminate all of the square roots and make it easier to solve.

But, I will give you a new method to solve this type of the equation.

a) \(\sqrt{x}+\sqrt{2-x}=2\) \(\left(0\le x\le2\right)\)

We can easily find that \(x=1\). When \(x=1\), \(\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{2-x}=1\end{matrix}\right.\). or \(\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{2-x}-1=0\end{matrix}\right.\) So, we have to do something like this:

\(\sqrt{x}+\sqrt{2-x}=2\Leftrightarrow\left(\sqrt{x}-1\right)+\left(\sqrt{2-x}-1\right)=0\) 

Notice that \(\sqrt{x}+1\ne0\) and \(\sqrt{2-x}+1\ne0\), we now can write the equation as below:

\(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\left(\sqrt{2-x}-1\right)\left(\sqrt{2-x}+1\right)}{\sqrt{2-x}+1}=0\) 

\(\Leftrightarrow\dfrac{\left(\sqrt{x}\right)^2-1}{\sqrt{x}+1}+\dfrac{\left(\sqrt{2-x}\right)^2-1}{\sqrt{2-x}+1}=0\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{1-x}{\sqrt{2-x}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{2-x}+1}\right)=0\)

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

\(\left(1\right)\Leftrightarrow\sqrt{x}=\sqrt{2-x}\Leftrightarrow x=1\)

Therefore, the equation a) has the root \(x=1\)

b)  \(0\le x\le1\)

Notice that \(x\) can be either equal to 0 or 1

So consider \(x=1\). Then, we have \(\sqrt{x}=1\Leftrightarrow\sqrt{x}-1=0\) and \(\sqrt{1-x}=0\). Therefore, we have to rewrite the equation like this:

\(\sqrt{1-x}+\sqrt{x}=1\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=0\)

\(\Leftrightarrow\dfrac{1-x}{\sqrt{1-x}}+\dfrac{x-1}{\sqrt{x}+1}=0\) \(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{1-x}}-\dfrac{1}{\sqrt{x}+1}\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\left(take\right)\\\dfrac{1}{\sqrt{1-x}}=\dfrac{1}{\sqrt{x}+1}\end{matrix}\right.\)or \(\left[{}\begin{matrix}x=1\\\sqrt{1-x}=\sqrt{x}+1\left(\cdot\right)\end{matrix}\right.\)

And now, use the same method to solve \(\left(\cdot\right)\)

c) We have \(x\ge0\)

We can easily see that \(x=4\), so \(\sqrt{x+5}=3\Leftrightarrow\sqrt{x+5}-3=0\) and \(\sqrt{x}=2\Leftrightarrow\sqrt{x}-2=0\) . Therefore, we can rewrite the equation as below:

\(\sqrt{x+5}-\sqrt{x}=1\Leftrightarrow\left(\sqrt{x+5}-3\right)-\left(\sqrt{x}-2\right)=0\) \(\Leftrightarrow\dfrac{\left(\sqrt{x+5}-3\right)\left(\sqrt{x+5}+3\right)}{\sqrt{x+5}+3}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+5}\right)^2-9}{\sqrt{x+5}+3}+\dfrac{\left(\sqrt{x}\right)^2-4}{\sqrt{x}+2}=0\)

\(\Leftrightarrow\dfrac{x-4}{\sqrt{x+5}+3}+\dfrac{x-4}{\sqrt{x}+2}=0\)

\(\Leftrightarrow\left(x-4\right)\left(\dfrac{1}{\sqrt{x+5}+3}+\dfrac{1}{\sqrt{x}+2}\right)=0\)

\(\Leftrightarrow...\)

Notice that \(\dfrac{1}{\sqrt{x+5}+3}+\dfrac{1}{\sqrt{x}+2}\) can't be equal to 0. So this equation only have the root \(x=4\)

d) Similar to the equations above.

Ý a, chỗ ( x-1 ) \(\dfrac{1}{\sqrt{x}+1}\) - \(\dfrac{1}{\sqrt{2-x}+1}\) = 0 tại sao lại làm mất được (x-1) vậy ạ ? 

a, \(x+x\cdot\dfrac{1}{3}:\dfrac{2}{9}+x:\dfrac{2}{7}=152\)

\(x+x\cdot\dfrac{1}{3}\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(x+x\cdot\dfrac{3}{2}+x\cdot\dfrac{7}{2}=252\)

\(x\left(1+\dfrac{3}{2}+\dfrac{7}{2}\right)=252\)

\(6x=252\)

\(\Rightarrow x=252:6=42\)

b, \(720:\left[41-\left(2\cdot x-5\right)\right]=18\)

\(\Rightarrow41-\left(2\cdot x-5\right)=720:18\)

\(41-\left(2x-5\right)=40\)

\(\Rightarrow2x-5=41-40\)

\(2x-5=1\)

\(2x=6\Rightarrow x=3\)

a)3/7x(-8/2)+(-3/5)= -2/7 + (-3/5)= -31/35

b)(4/3)+(-2/5)+(-3/2)= 14/15 + (-3/2)= -17/30

c)4/5-(-2/7)- 7/10 =38/35 - 7/10 = 27/70

\(\dfrac{9^9+27^7}{9^6+243^3}=\dfrac{\left(3^2\right)^9+\left(3^3\right)^7}{\left(3^2\right)^6+\left(3^5\right)^3}=\dfrac{3^{18}+3^{21}}{3^{12}+3^{15}}=\dfrac{3^{18}\left(1+3^3\right)}{3^{12}\left(1+3^3\right)}\) \(=\dfrac{3^{18}}{3^{12}}=3^6\)

(-3/4)^2= 9/16

\((-\dfrac{3}{4})^2=\dfrac{9}{16}\)