a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)

b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

c, \(\left(x-3\right)\left(2x-5\right)=0\)

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

d, \(\left(x-3\right)x-2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)

\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

g, \(x^2+6x-7=0\)

\(\Rightarrow x^2-x+7x-7=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

h,\(2x^2+5x-7=0\)

\(\Rightarrow2x^2-2x+7x-7=0\)

\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Chúc bạn học tốt!!!

a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)

b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

vậy \(x=8;x=-2\)

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

vậy \(x=3;x=\dfrac{5}{2}\)

d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)

e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)

câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha

g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)

h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)

\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)

Ta có : \(\left|2x+4\right|+\left|4x+8\right|=0\left|2x+4\right|+\left|4x+8\right|=0\)

\(\Rightarrow\left|2x+4\right|+2.\left|2x+4\right|=\left|4x+8\right|=0\)

\(\Rightarrow\left|2x+4\right|\left(1+2\right)=0\)

=> |2x + 4| = 0

=> 2x + 4 = 0

=> 2x = -4

=> x = -2

1. Đề đúng phải là thế này: \(\left|2x+4\right|+\left|4x+8\right|=0\)

\(\Rightarrow\left|2x+4\right|=\left|4x+8\right|=0\)

\(\Rightarrow2x+4=4x+8=0\)

\(\Rightarrow x=-\frac{4}{2}=-\frac{8}{4}\)

\(\Rightarrow x=-2\)

2. Sửa lại đề : \(\left|x-5\right|-\left|x-7\right|=0\)

\(\Rightarrow\left|x-5\right|=\left|x-7\right|\)

\(\Rightarrow\orbr{\begin{cases}x-5=x-7\\x-5=-\left(x-7\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-5=-7\\x-5=-x+7\end{cases}}\)

( Loại trường hợp 1)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\)

3.  \(\left|x+8\right|-\left|2x+2\right|=0\)

\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)

\(\Rightarrow\orbr{\begin{cases}x+8=2x+2\\x+8=-\left(2x+2\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+8=-2x-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\3x=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)

Trả lời

Mk nghĩ bạn có thể tham khảo ở CHTT nha !

Có đáp án của câu b;c và d đó.

Đừng ném đá chọi gạch nha !

a) vi(x^2+5)(x^2-25)=0

=>x^2+5=0 hoac x^2-25=0

=>x=...hoac x=...(tu lam)

b)(x-2)(x+1)=0

=>x-2=0 hoac x+1=0

=>x=2 hoac x=-1

c)(x^2+7)(x^2-49)<0

=>x^2+7va x^2-49 trai dau

ma x^2+7>=7=>x^2-49<0=>x<7 va x>-7

con lai tuong tu

tu lam nhe nho k nha

bài 2) a) \(2\left(x+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\) vậy \(x=-1\)

b) \(x\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)

c) \(\left(x-1\right)\left(x+7\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) vậy \(x=1;x=-7\)

d) \(\left(x+2\right)\left(x^2-9\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x^2-9=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x^2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\end{matrix}\right.\) vậy \(x=-2;x=3;x=-3\)

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

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

bài 1) \(A=48+\left(-48-174\right)+\left|-74\right|=48-48-174+74=-100\)

\(B=\left(-123\right)+77+\left(-257\right)-23-43=-123+77-257-23-43=-369\)

\(C=\left(-57\right)+\left(-159\right)+47+169=-57-159+47+169=0\)

quá hợp lí

999 - 888 - 111 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111 + 111 - 111

= 0 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111

= 0 + 111 - 111

= 111 - 111

= 0

999 - 888 - 111 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111 + 111 - 111

= 0 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111

= 0 + 111 - 111

= 111 - 111

= 0

1) -12(x-5)+7(3-x)=5

=> (-12x)-(-60)+21-7x=5

=> (-12x)+60+21-7x  =5

=> (-12x)+81-7x        =5

=> (-12x)-7x              =5-81

=> x.[(-12)-7)]            =-76

=> x.(-19)                  =-76

=> x                          =(-76):(-19)

=> x                          =4

Vậy x=4

2) (x-2).(x+4)=0

Để tích trên bằng 0 <=> x-2=0=>x=0+2=>x=2 và x+4=0=>x=0-4=>x=-4

Vậy x =-4;2

3) (7-x).(x+19)=0

Để tích trên bằng 0 <=> 7-x=0=>x=7-0=>x=7 và x+19=0=>x=0-19=>x=-19

Vậy x=-19;7

4) (x-2).(x+15)=0

Để tích trên bằng 0 <=> x-2=0=>x=0=2=>x=2 và x+15=0=>x=0-15=>x=-15

Vậy x=-15;2

có gì đó sai sai________@_@

không phải đâu bạn ơi. Bài này bạn phải xét 3 trường hợp là x = 0 ; (x-7) = 0 và (x-3) = 0 cơ.