\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

1.

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow3x^2-3x+\left(x+1-\sqrt{3x+1}\right)+\left(x+2-\sqrt{5x+4}\right)=0\)

\(\Leftrightarrow3\left(x^2-x\right)+\dfrac{x^2-x}{x+1+\sqrt{3x+1}}+\dfrac{x^2-x}{x+2+\sqrt{5x+4}}=0\)

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

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

\(\Leftrightarrow...\)

2.

Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt[3]{2-8x^3}=b\end{matrix}\right.\)

Ta được hệ:

\(\left\{{}\begin{matrix}\left(2a-1\right)b=a\\a^3+b^3=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2ab\\\left(a+b\right)^3-3ab\left(a+b\right)=2\end{matrix}\right.\)

\(\Rightarrow8\left(ab\right)^3-6\left(ab\right)^2=2\)

\(\Leftrightarrow\left(ab-1\right)\left[4\left(ab\right)^2+ab+1\right]=0\)

\(\Leftrightarrow ab=1\Rightarrow a+b=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\ab=1\end{matrix}\right.\) \(\Leftrightarrow a=b=1\)

\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

a) 16 - 3x = 4

<=> 3x = 12

<=> x = 4

Vậy x = 4 là nghiệm phương trình 

b) (x² - 4x + 5)² - (x - 1)(x - 3) = 4

<=> (x² - 4x + 5)² - 4 - (x - 1)(x - 3) = 0

<=> (x² - 4x + 5 - 2)(x² - 4x + 5 + 2) - (x - 1)(x - 3) = 0

<=> (x² - 4x + 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 6) = 0

<=> (x  - 1)(x - 3) = 0 (Vì x² - 4x + 6 > 0 \(\forall x\))

<=> \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy x \(\in\left\{1;3\right\}\)là nghiệm phương trình 

a)16-3x=4

3x=16-4

3x=12

x=4

Vậy x=4

b)(x²-4x+5)²-(x-1).(x-3)=4

[(x-2)²+1]²-[(x-2)+1].[(x-2)-1]=4

=>(x-2)²+2.(x-2).1+1-(x-2)²-1²=4

2(x-2)=4

=>x-2=2

=>x=4

Vậy ....................

Chú bn học tốt

a) 16 - 3x = 4

<=> 3x = 12

<=> x = 4

Vậy x = 4 là nghiệm phương trình 

b) (x² - 4x + 5)² - (x - 1)(x - 3) = 4

<=> (x² - 4x + 5)² - 4 - (x - 1)(x - 3) = 0

<=> (x² - 4x + 5 - 2)(x² - 4x + 5 + 2) - (x - 1)(x - 3) = 0

<=> (x² - 4x + 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 6) = 0

<=> (x  - 1)(x - 3) = 0 (Vì x² - 4x + 6 > 0 ∀x)

<=> [

⇔[

Vậy x ∈{1;3}là nghiệm phương trình 

khó quá ?????

mik chưa học đến lớp 8 nên ko biết

khó thể xem trên mạng

bài 1 câu a bỏ x= nhé !

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-3b^2-2ab=0\)

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

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)

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

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

a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )  

<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0

<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0

<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0

<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)

Vậy x = { \(\frac{-1}{3};-5\)} 

b) x² + 10.x + 25 - 4.x . ( x + 5 ) = 0 

<=> ( x + 5 )² -4.x . (x + 5 ) = 0

<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0

<=> ( x + 5 ) . ( 5 - 3.x )  = 0

<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{3};-5\right\}\)

c) (4.x - 5 )² - 2. ( 16.x² -25 ) = 0 

<=> ( 4.x-5)² -2 .( 4.x-5) .( 4.x + 5 ) = 0

<=> (  4.x -5 )² - ( 8.x+ 10 ) . ( 4.x -5 ) = 0

<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0

<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0

<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)

d) ( 4.x + 3 )² = 4. ( x² - 2.x + 1 ) 

<=> 16.x² + 24.x + 9 - 4.x²  + 8.x - 4 = 0

<=> 12.x² + 32.x + 5 =0 

<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0 

<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

e) x² -11.x + 28 = 0

<=> x² -4.x  - 7.x + 28 = 0

<=> ( x - 7 ) . ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy x = { 4 ; 7 } 

f ) 3.x.3 - 3.x² - 6.x = 0

<=> 3.x. ( x² -x - 2 ) = 0 

<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

        \([x=0\)                \([x=0\)

( Lưu ý :Lưu ý này không cần ghi vào vở :  Chị nối 2 ý đó làm 1 nha cj ! ) 

Vậy x = { 2 ; -1 ; 0 }