2. you may assign a room for me?

1)I suggested that he buy a alarm clock, because of the fact that never does he come to class on time.

-> I suggested that he should buy/buy an alarm clock, because of the fact that he never come to class on time.

2)các anh có thể NHƯỜNG lại phòng cho tôi được không ?

-> Can you (boys) make/give the room for me ?

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

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

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

Câu 1 :

a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

b. 

|-2x|=5x+14

 Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14

 <=> - 2x = 5x + 14

 <=> - 2x - 5x = 14

 <=> - 7x = 14

 <=> x = - 2 (thoã mãn)

 Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.

Ta có pt: 2x = 5x + 14

 <=> - 3x = 14

<=> x = \(-\dfrac{14}{3}\)
 Vậy pt có nghiệm x = - 2

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)

\(a,4x-5=23\)

\(\Leftrightarrow4x=23+5\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(b,\left|-2x\right|=5x+14\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)

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

\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

a) \(4x-5=23\)

    \(4x=23+5\)

      \(4x=28\)

        \(x=7\)

b) \(\left|-2x\right|=5x+14\)

 \(\Leftrightarrow\)   \(-2x-5=14\)

\(\Leftrightarrow\)    \(-7x=14\)

\(\Leftrightarrow\)         \(x=-2\)

\(\Leftrightarrow\)    \(-2x=-\left(5x+14\right)\)

\(\Leftrightarrow\)    \(-2x=-\left(5x-14\right)\)

\(\Leftrightarrow\)  \(-2x+5x=-14\)

 \(\Leftrightarrow\)    \(3x=-14\)

 \(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)

  \(\Leftrightarrow x=-2\)   

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)

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

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

\(\Leftrightarrow x^2+x+2=x^2+2\)

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)

dễ mk bn cho mình hỏi nhé câu 4 là \(\frac{1}{2\cdot3}\)hay là\(\frac{1}{2}\cdot3\)

Cau 1

A=17/72     B=16      C=-5/21

Cau 2

a)    x=1      b)    -10/3

a) = (x + 3)² - y² = (x + 3 - y)(x + 3 + y)

b) = x²(x - 3) -4(x - 3) = (x - 3)(x² - 4) = (x - 3)(x - 2)(x + 2)

c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)

d) Nhầm đề. tui sửa lại x³ + y³ + 2x² - 2xy + 2y²

= x³ + y³ + 2(x² - xy + y²) = (x + y)(x² - xy + y²) + 2(x² - xy + y²) = (x² - xy + y²)(x + y + 2)

e) = x⁴ - x³ - x³ + x² - x² + x + x - 1 = x³(x - 1) - x²(x - 1) - x(x - 1) + x - 1 = (x - 1)(x³ - x² - x + 1) = (x - 1)(x - 1)(x² - 1) = (x - 1)³(x + 1)

f) = x³ - 3x² - x² + 3x + 9x - 27 = x²(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x² - x + 9)

g) chắc là 3xyz 

= x²y + xy² + y²z + yz² + x²z + xz² + 3xyz = x²y + xy² + xyz + y²z + yz² + xyz + x²z + xz² + xyz = (x + y + z)(xy + yz + xz)

h) = 2³ -(3x)³ = (2 - 3x)(4 + 6x + 9x²)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x³ - y³)(x³ + y³) = (x - y)(x² + xy +y2)(x + y)(x² - xy +y2).

4:

a: 7/10

b: 6/5=12/10

7/2=35/10

8/25=32/100

11/50=22/100

23/125=204/1000

17/20=85/100

41/200=205/1000

đề là gì? 

mk cần gấp

x².( x² + 4 ) - x² - 4

= x².( x² + 4 ) - ( x²  + 4 )

= ( x² + 4 ).( x² - 1 )

= ( x² + 4 ) .( x - 1 ).( x + 1 )

TL:

\(x^2\left(x^2+4\right)-x^2-4\) 

\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\) 

\(=\left(x^2-1\right)\left(x^2+4\right)\)