\(\Rightarrow\dfrac{1}{\left(2x-1\right)\left(2x-2\right)}+\dfrac{1}{\left(2x-2\right)\left(2x-3\right)}+\dfrac{1}{\left(2x-3\right)\left(2x-4\right)}+\dfrac{1}{\left(2x-4\right)\left(2x-5\right)}=\dfrac{4}{21}\)

\(\Rightarrow\dfrac{1}{2x-2}-\dfrac{1}{2x-1}+\dfrac{1}{2x-3}-\dfrac{1}{2x-2}+\dfrac{1}{2x-4}-\dfrac{1}{2x-3}+\dfrac{1}{2x-5}-\dfrac{1}{2x-4}=\dfrac{4}{21}\)

\(\Rightarrow\dfrac{1}{2x-5}-\dfrac{1}{2x-1}=\dfrac{4}{21}\)

\(\Rightarrow\dfrac{4}{\left(2x-1\right)\left(2x-5\right)}=\dfrac{4}{21}\)

\(\Rightarrow\left(2x-1\right)\left(2x-5\right)=21\)

\(\Rightarrow4x^2-12x-16=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

\(\dfrac{x+2}{x-1}=\dfrac{x-1}{x-3}\) (1)

ĐKXĐ: \(x\ne1;x\ne3\)

(1) \(\Leftrightarrow\left(x+2\right)\left(x-3\right)=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-3x+2x-6=x^2-2x+1\)

\(\Leftrightarrow-3x+2x+2x=1+6\)

\(\Leftrightarrow x=7\) (nhận)

Vậy S = {7}

Bài 1: 

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

Ta có: \(\sqrt{5x^2}=2x-1\)

\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)

\(\Leftrightarrow5x^2-4x^2+4x-1=0\)

\(\Leftrightarrow x^2+4x-1=0\)

\(\text{Δ}=4^2-4\cdot1\cdot\left(-1\right)=20\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-4-2\sqrt{5}}{2}=-2-\sqrt{5}\left(loại\right)\\x_2=\dfrac{-4+2\sqrt{5}}{2}=-2+\sqrt{5}\left(loại\right)\end{matrix}\right.\)

Bài 1: Bình phương hai vế lên có giải ra được kết quả. Nhưng phải kèm thêm điều kiện $2x-1\geq 0$ do $\sqrt{5x^2}\geq 0$

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 5x^2=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x^2+4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ (x+2)^2-5=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ (x+2-\sqrt{5})(x+2+\sqrt{5})=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x=-2\pm \sqrt{5}\end{matrix}\right.\) (vô lý)

Vậy pt vô nghiệm.

Bài 2: ĐKXĐ luôn là thứ mà phải ghi ngay đầu bài làm để xác định được biểu thức có nghĩa. Tức là em ghi ĐKXĐ: $x+1\geq 0$ đầu tiên.

Sau đó mới giải ra $\sqrt{x+1}=1$

1 have waited

2 has written

3 has studied

4 has finished

5 haven't done

6 We don't go to school on Saturdays.

7 They usually hold a party in New Year's Eve.

8 I've seen this film already.

9 Last night, she came home between 10.30 and 11 o'clock.

10 We are going to see the movie Dream City at 7.00 this evening.

1. have waited

2. has written

3. has studied

4. has finished

5 haven't done

6. We don't go to school on Saturdays

7. They usually hold a party on New Year's Eve

8. I have seen this film already

9. Last night she came home between 10.30 and 11 o'clock

10. We are going to see the movie Dream City at 7.00 this evening

\(x^2\left(x+4,5\right)=13,5\)

<=>\(x^3+4,5x^2-13,5=0\)

<=> \(x^3+3x^2+1,5x^2+4,5x-4,5x-13,5=0\)

<=>\(x^2\left(x+3\right)+1,5x\left(x+3\right)-4,5\left(x+3\right)=0\)

<=>\(\left(x+3\right)\left(x^2+1,5x-4,5\right)=0\)

<=>\(\left(x+3\right)\left[x^2+3x-1,5-4,5\right]=0\)

<=>\(\left(x+3\right)\left[x\left(x+3\right)-1,5\left(x+3\right)\right]=0\)

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

<=> \(\left[{}\begin{matrix}\left(x+3\right)^2=0\\x-1,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1,5\end{matrix}\right.\)

Vậy...

Ta có: \(x^2\left(x+4.5\right)=13.5\)

\(\Leftrightarrow x^3+\dfrac{9}{2}x^2-\dfrac{27}{2}=0\)

\(\Leftrightarrow2x^3+9x^2-27=0\)

\(\Leftrightarrow2x^3-3x^2+12x^2-18x+18x-27=0\)

\(\Leftrightarrow x^2\left(2x-3\right)+12x\left(2x-3\right)+9\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x^2+12x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+12x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\\left(x+6\right)^2=27\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x+6=3\sqrt{3}\\x+6=-3\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\sqrt{3}-6\\x=-3\sqrt{3}-6\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};3\sqrt{3}-6;-3\sqrt{3}-6\right\}\)

Help me

x³ - 6xy + y³ = 8

<=> (x + y)³ - 3xy(x + y) - 6xy + 8 = 16

<=> (x + y + 2)(x² + y² - xy - 2x - 2y + 4) = 16

<=> \(\left(x+y+2\right)\left[\left(x-\dfrac{1}{2}y-1\right)^2+3\left(\dfrac{1}{2}y-1\right)^2\right]=16\)

Nhận thấy \(\left(x-\dfrac{1}{2}y-1\right)^2+3\left(\dfrac{1}{2}y-1\right)^2\ge0\)

=> x + y + 2 > 0

Khi đó 16 = 1.16 = 2.8 = 4.4

Lập bảng 

 Đến đó bạn thế x qua y rồi làm tiếp nha

Bài 8:

a: Khi a=1 thì phương trình sẽ là \(\left(1-4\right)x-12x+7=0\)

=>-3x-12x+7=0

=>-15x+7=0

=>-15x=-7

hay x=7/15

b: Thay x=1 vào pt, ta được:

\(a^2-4-12+7=0\)

\(\Leftrightarrow\left(a-3\right)\left(a+3\right)=0\)

hay \(a\in\left\{3;-3\right\}\)

c: Pt suy ra là \(\left(a^2-16\right)x+7=0\)

Để phương trình đã cho luôn có một nghiệm duy nhất thì (a-4)(a+4)<>0

hay \(a\notin\left\{4;-4\right\}\)