`@` `\text {Ans}`

`\downarrow`

`a)`

\(1\dfrac{3}{5}\div\left(-5\dfrac{5}{7}\right)\)

`=`\(\dfrac{8}{3}\div\left(-\dfrac{30}{7}\right)\)

`=`\(\dfrac{8}{3}\cdot\left(-\dfrac{7}{30}\right)=-\dfrac{28}{45}\)

`b)`

\(-1\dfrac{1}{8}\cdot\dfrac{4}{51}\cdot\left(-11\dfrac{1}{3}\right)=-\dfrac{9}{8}\cdot\dfrac{4}{51}\cdot\left(-\dfrac{34}{3}\right)=-\dfrac{3}{34}\cdot\left(-\dfrac{34}{3}\right)=1\)

a)...\(\dfrac{8}{5}:\left(-\dfrac{40}{7}\right)=\dfrac{8}{5}.\left(-\dfrac{7}{40}\right)=-\dfrac{7}{25}\)

b) \(...-\dfrac{9}{8}.\dfrac{4}{51}\left(-\dfrac{34}{3}\right)=1\)

\(\sqrt[]{x+3}+\sqrt[]{x-1}=2\left(x\ge1\right)\)

\(\Leftrightarrow x+3+x-1+2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4\)

\(\Leftrightarrow2x+2+2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4\)

\(\Leftrightarrow2\sqrt[]{\left(x+3\right)\left(x-1\right)}=4-2\left(x+1\right)\)

\(\Leftrightarrow\sqrt[]{\left(x+3\right)\left(x-1\right)}=2-\left(x+1\right)\)

\(\Leftrightarrow\sqrt[]{\left(x+3\right)\left(x-1\right)}=1-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\\Leftrightarrow\left(x+3\right)\left(x-1\right)=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow x^2+2x-3=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\Leftrightarrow x=1\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

Điều kiện xác định: \(x\ge1\)

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

\(\Leftrightarrow x+1+\sqrt{x^2+2x-3}=2\\\Leftrightarrow\sqrt{x^2+2x-3}=1-x \)

Để phương trình thỏa mãn thì x\(\le1\)mà \(x\le1\)

\(\Rightarrow x=1\)

Thử lại, ta được: \(\sqrt{1+3}+\sqrt{1-1}=2\left(tm\right)\)

Vậy x=1

a) \(...-\dfrac{22}{7}:-\dfrac{15}{9}=\dfrac{22}{7}.\dfrac{9}{15}=\dfrac{66}{35}\)

b) \(...\dfrac{8}{3}:\left(-\dfrac{42}{7}\right)=\dfrac{8}{3}.\left(-\dfrac{7}{42}\right)=-\dfrac{8}{9}\)

c) ...\(\left(-\dfrac{7}{2}\right):\left(-\dfrac{13}{5}\right)=\dfrac{7}{2}.\dfrac{5}{13}=\dfrac{35}{36}\)

\(\dfrac{x}{y}=3;\dfrac{y}{z}=2\Rightarrow\dfrac{x}{y}.\dfrac{y}{z}=6\Rightarrow\dfrac{x}{z}=6\)

`@` `\text {Ans}`

`\downarrow`

Ta có:

`x` tỉ lệ thuận với `y` theo hệ số tỉ lệ `k = 3`

`=> x = 3y` `(1)`

`y` tỉ lệ thuận với `z` theo hệ số tỉ lệ `k = 2`

`=> y = 2z` `(2)`

Thay `(2)` vào `(1)`

`x = 3*2*z`

`=> x = 6*z`

Vậy, `x` tỉ lệ thuận với `z` theo hệ số tỉ lệ `6.`

Bạn xem lại đề nhé, mình chưa hiểu lắm 

\(-\dfrac{12}{21}\div\dfrac{34}{43}\)

\(=-\dfrac{12}{21}\times\dfrac{43}{34}\)

\(=-\dfrac{86}{119}\)

\(-\dfrac{12}{21}:\dfrac{34}{43}=-\dfrac{12}{21}.\dfrac{43}{34}=-\dfrac{2}{7}.\dfrac{43}{17}=-\dfrac{86}{119}\)

\(\dfrac{17}{15}\div\dfrac{4}{3}\)

\(=\dfrac{17}{15}\times\dfrac{3}{4}\)

\(=\dfrac{17}{20}\)

\(4x\left(7x^3-6x^2-8x\right)\)

\(=28x^4-24x^3-32x^2\)

Lời giải:

1. Dấu giữa (x+3) và (2x+3)² là gì vậy bạn?

2.

$E=(4x^2-12x)-(x^2-10x+25)-3(x+1)^2+4(x+1)^2-4x^2+5$

$=4x^2-12x-x^2+10x-25+(x+1)^2-4x^2+5$

$=4x^2-12x-x^2+10x-25+x^2+2x+1-4x^2+5$

$=(4x^2-x^2+x^2-4x^2)+(-12x+10x+2x)+(-25+1+5)$

$=-19$ là giá trị không phụ thuộc vào biến (đpcm)

    13 \(\times\) 52 + 52 \(\times\) 35 - 52 \(\times\) 19

= 52 \(\times\) ( 13 + 35 - 19)

= 52 \(\times\) 29

= 1508