a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)

\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)

Vậy pt vô nghiệm 

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

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)

c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)

Trả lời:

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\)

\(\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)

\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)

\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)

Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)

nên pt vô nghiệm 

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

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)

\(\Leftrightarrow3x.\left(-9\right).2x=0\)

\(\Leftrightarrow-54x^2=0\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt.

c, \(7-9x+2x^2=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)

Vậy x = 7/2; x = 1 là nghiệm của pt.

d, trùng ý c

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)

Bài 5 : 

a, \(2x\left(x-3\right)+x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(x\left(x+1\right)-x-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=\pm1\)

c, sửa đề  \(x^3-3x^2+x-3=0\Leftrightarrow x^2\left(x-3\right)+x-3=0\)

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

d, \(3x^2\left(2x-1\right)+1-4x^2=0\Leftrightarrow3x^2\left(2x-1\right)+\left(1-2x\right)\left(1+2x\right)=0\)

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

e, \(x^3+2x-x^2-2=0\Leftrightarrow x\left(x^2+2\right)-\left(x^2+2\right)=0\)

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

x=1 nha

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

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Bài 5 : 

f, bạn xem lại đề hay là tìm x chứa tham số a ? 

g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

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

h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)

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

m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)

n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)

x=7 nha

a)x²-3x+2=0

=>x²-2x-x+2=0

=>x.(x-2)-(x-2)=0

=>(x-2)(x-1)=0

=>x-2=0 hoặc x-1=0

=>x=2 hoặc x=1

b)2x²-5x+3<0

=>2x²-2x-3x+3<0

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

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

TH1: x-1 >0 và 2x-3<0

=>x>1 và x<3/2

=>1<x<3/2

TH2: x-1<0 và 2x-3>0

=>x<1 và x>3/2(vô lí)

Vậy 1<x<3/2

còn câu c bạn tự giải nha

a)\(\left(4-x\right)^2-16=0\)

\(\Leftrightarrow\left(4-x\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)

b) \(25-\left(3-x\right)^2=0\)

\(\Leftrightarrow\left(3-x\right)^2=25\)

\(\Leftrightarrow\orbr{\begin{cases}3-x=5\\3-x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

c)\(3x^2-6x+3-27=0\)

\(\Leftrightarrow3x^2-6x-24=0\)

\(\Leftrightarrow\left(3x+6\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}}\)

#H

1.(4-x)²-16 =0
<=> 16 -8x+x² -16 =0
<=> -x(8-x) =0
<=> TH1: x=0
.      TH2: 8-x=0
.              => x= -8

2. 25 - (3-x)² = 0
<=> 25 - (9-6x+x²) = 0
<=> 25 - 9+6x-x² = 0
<=> -x²+6x+16 = 0
<=> -(x-8)(x+2) = 0 (bước này bạn nhập phương trình trên mtinh là nó sẽ ra nghiệm nhe)
<=> TH1:x-8=0
.             x= 8
.      TH2:x+2=0
.             x=-2

3.(bạn tự làm nhé, giải bth thui)

Trả lời:

1, \(\left(4-x\right)^2-16=0\)

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

\(\Leftrightarrow\left(-x\right)\left(8-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\8-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)

Vậy x = 0; x = 8 là nghiệm của pt.

2, \(25-\left(3-x\right)^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2+x=0\\8-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

Vậy x = - 2; x = 8 là nghiệm của pt.

3, \(3x^2-6x+3-27=0\)

\(\Leftrightarrow3x^2-6x-24=0\)

\(\Leftrightarrow3\left(x^2-2x-8\right)=0\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow x^2-4x+2x-8=0\)

\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)

Vậy x = 4; x = - 2 là nghiệm của pt.

Trả lời:

\(1,3x\left(x-7\right)+2x-14=0\)

\(\Leftrightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{2}{3}\end{cases}}}\)

Vậy x = 7; x = - 2/3 là nghiệm của pt.

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}}\)

Vậy x = - 3; x = 1; x = - 1 là nghiệm của pt.

\(3,15x-5+6x^2-2x=0\)

\(\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

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

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

Vậy x = 1/3; x = - 5/2 là nghiệm của pt.