1 ) \(\left(x-4\right)^2-25=0\)

\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow-2\left(2x-4\right)=0\)

\(\Leftrightarrow x=2.\)

3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)

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

5 ) \(x^3+x^2+x+1=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

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

6 ) \(x^3+x^2-x-1=0\)

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

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

7 ) \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=-1.\)

8 ) \(x^4-4x^3-19x^2+106x-120=0\)

\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)

\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)

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

9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)

\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)

Đặt \(x^2+6x-7=t\)

\(\Leftrightarrow t\left(t-9\right)+8=0\)

\(\Leftrightarrow t^2-9t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)

Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)

Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)

Vậy ........

1,

\(x^2+y^2+y^2=14\)

\(\Rightarrow\left(x+y+z\right)^2-2xy-2yz-2zx=14\)

\(\Rightarrow-2\left(xy+yz+zx\right)=14\)

\(\Rightarrow xy+yz+zx=-7\)

\(\Rightarrow\left(xy+yz+zx\right)^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=49\)

Ta có: \(x^4+y^4+z^4\)

\(=\left(x^2+y^2+z^2\right)^2-2x^2y^2-2y^2z^2-2z^2x^2\)

\(=14^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=14^2-2.49\)

\(=196-98\)

\(=98\)

Bài 2: 

a: x(x-3)<0

=>x>0 và x-3<0

=>0<x<3

mà x là số nguyên

nên \(x\in\left\{1;2\right\}\)

b: x(x+2)<0

=>x+2>0và x<0

=>-2<x<0

mà x là số nguyên

nen x=-1

c: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

\(\Leftrightarrow1< x^2< 4\)

mà x là số nguyên

nên \(x\in\varnothing\)

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

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

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

=>-19x+81=5

=>-19x=5-81

=>-19x=-76

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

=>x=4

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

=>+,x-2=0 => x=2

+,x+4=0 => x=-4

Vậy x=2 hoặc x=-4

3,(x-2) (x+15) =0

=>+,x-2=0 =>x=2

+,x+15=0 =>x=-15

Vậy x=2 hoặc x=-15

4,(7-x) (x+19) =0

=>+,7-x=0 =>x=7

+,x+19=0 =>x=-19

Vậy x=7 hoặc x=-19

5,(x-3) (x-5)<0

=>x-3 và x-5 là hai số khác dấu

TH1

+,x-3<0 =>x<3(1)

+,x-5>0 =>x>5 (2)

Từ (1) và(2) => 5<x<3(Vô lí nên trường hợp này bị loại)

TH2

+,x-3>0 =>x>3 (3)

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

Từ (3) và (4) =>3<x<5 => x=4

Vậy x=4

Chú bn hc tốt hơn nha!!

hình như sai đề câu b vs d bn ơi

x là nhân ak

khó quá

xin lỗi nhé mik ko làm đc

nhìn hoa mắt và nhiều quá

Chúc bạn học tốt!

a)      \(4.\left(x+5\right)< 0\)

\(\Rightarrow\)\(4\) và     \(x+5\)   trái dấu

mà    \(4>0\)

nên       \(x+5< 0\)

\(\Rightarrow\)\(x< -5\)

de qua

Bài 2: 

a: \(x^3-\dfrac{1}{4}x=0\)

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

hay \(x\in\left\{0;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

b: \(x^2-10x=-25\)

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

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

=>x-5=0

hay x=5

c: \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

hay \(x\in\left\{0;-\sqrt{13};\sqrt{13}\right\}\)

d: \(x^2+2x-1=0\)

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

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

hay \(x\in\left\{\sqrt{2}-1;-\sqrt{2}-1\right\}\)