b) Ta có: \(\left(5-x\right)^3+27=0\)

\(\Leftrightarrow\left(5-x\right)^3=-27\)

\(\Leftrightarrow5-x=-3\)

hay x=8

d) Ta có: \(\left(x^2-1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-7\end{matrix}\right.\)

f) Ta có: \(\left(x^2+81\right)\left(x-7\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> ( 2x - 1 )² + ( x + 3 )² - 5( x² - 7² ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> x = -255/2

\(pt< =>4x^2-4x+1+x^2+6x+9-5x^2+5.49=0\)

\(< =>2x+255=0< =>x=-\frac{255}{2}\)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x+255=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=\frac{-255}{2}\)

Vậy tập nghiệm của phương trình là: \(S=\left\{\frac{-255}{2}\right\}\)

bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35

Mk ko chép lại đề nha !

\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(2x+255=0\)

\(2x=-255\)

\(x=\frac{-225}{2}\)

(2x - 1)² + (x + 3)² - 5(x + 7)(x - 7)      = 0

4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

                                             2x + 255 = 0

                                             2x           = -255

                                               x           = -127,5.

\(\left(x-5\right)^2-7\left(5-x\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2+7\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)+7\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

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

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

Vậy \(S=\left\{5;-2\right\}\)

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

Chỗ dấu "..." bạn không cần ghi.Mình viết vậy cho dễ nhìn. Bài này có một lời giải khá độc đáo trong sách nâng cao của mình.

a) Số thừa số âm ở VT chẵn.

Mà \(x-\frac{2}{5}< x+\frac{3}{7}< x+\frac{3}{4}\)  nên

\(\orbr{\begin{cases}x-\frac{2}{5}>0\\x+\frac{3}{7}< 0..và...x+\frac{3}{4}>0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x>\frac{2}{5}\\x< -\frac{3}{7}...và...x>-\frac{3}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x>\frac{2}{5}\\-\frac{3}{4}< x< -\frac{3}{7}\end{cases}}}\)

len google di ban

mk chua hoc bai nay

a,\((x+4)^2-(x+1)(x-1)=16\)

 \(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)

b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)

\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)

\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)

\(\Rightarrow -x^2-10x-244=0\)

\(\Rightarrow -(x^2-10x+25)-219=0\)

\(\Rightarrow -(x-5)^2-219=0\)

\(\Rightarrow (x-5)^2+219=0\)

Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm