Đề bạn có mấy chỗ thiếu mk bổ sung nha

\(a,2^3+4^2+6x=8+16+6x=6x+24=x\left(x+4\right)\\ b,x^2-4=\left(x-2\right)\left(x+2\right)\\ c,x^2-10x+25=\left(x-5\right)^2\\ d,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ e,x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ g,x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

Tick plzz

a: Ta có: \(2x^3+4x^2+6x\)

\(=2x\left(x^2+2x+3\right)\)

b: \(x^2-4=\left(x-2\right)\left(x+2\right)\)

c: \(x^2-10x+25=\left(x-5\right)^2\)

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

e: \(x^2+xy-3x-3y\)

\(=x\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x-3\right)\)

g: \(x^2-4x+4-y^2\)

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

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

Câu 1:

\(2x^3-3x^2+x+a\)

\(=2\left(x^3-6x^2+12x-8\right)+9\left(x^2-4x+4\right)+13\left(x-2\right)+\left(6+a\right)\)

\(=2\left(x-2\right)^3+9\left(x-2\right)^2+13\left(x-2\right)+\left(6+a\right)\)chia hết cho \(x-2\)khi và chỉ khi :

\(6+a=0\Leftrightarrow a=-6\). Vậy \(a=-6\).

Câu 2:

\(\left(x+1\right)\left(2x-x\right)-\left(3x+5\right)\left(x+2\right)=4x^2+1\)

\(\Leftrightarrow x^2+x-\left(3x^2+11x+10\right)=-4x^2+1\)

\(\Leftrightarrow x^2+x-3x^2-11x-10+4x^2-1=0\)

\(\Leftrightarrow2x^2-10x-11=0\)

\(\Delta'=\left(-5\right)^2-2\left(-11\right)=47>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt:

\(x=\frac{5+\sqrt{47}}{2}\)hoặc \(x=\frac{5-\sqrt{47}}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{5+\sqrt{47}}{2};\frac{5-\sqrt{47}}{2}\right\}\)

=>(x+m)(x-1)+x^2-9=2(x^2+2x-3)

=>x^2-x+mx-m+x^2-9=2x^2+4x-6

=>x(m-5)=-6+m+9=m+3

Để phương trình có nghiệm duy nhất thì m-5<>0

=>m<>5

bài 2 là tìm X nha mn

\(1,\\ a,=x\left(2x+3y-5\right)\\ b,=x\left(x-2y\right)+\left(x-2y\right)=\left(x+1\right)\left(x-2y\right)\\ 2,\\ a,\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow x\left(x-2y\right)+\left(x-2y\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2y\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2y\left(y\in R\right)\end{matrix}\right.\)

Mik quên là bài này là bài tìm x nha mọi người ❤

a)( x+ 5)+(x-9)=x+2

x + 5 + x - 9 = x + 2

x + x - x = 2 -5 +9

x = 6

b)( 27 -x) +( 15+x )=x-24

-x + x - x = -24 -27 -15

-x = -66

x=66

hok tốt!

\(a.\left(x+5\right)+\left(x-9\right)=\)\(x+2\)

\(\Rightarrow x+x-x=2+9-5\)

\(\Rightarrow x=6\)

\(b.\left(27-x\right)+\left(15+x\right)=x-24\)

\(\Rightarrow x-x-x=-24-27-15\)

\(\Rightarrow-x=-66\)

\(\Rightarrow x=66\)

(x+5)+(x-9)=x+2

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

<=> x+5+x-9-x-2=0

<=> x-6=0

<=> x=6

Vậy x=6

(27-x)+(15+x)=x-24

<=> 27-x+15+x=x-24

<=> 27-x+15+x-x+24=0

<=>66-x=0

<=> x=66

Vậy x=66

a) x vô nghĩa

b) x=0,8;x=-0,8

d: Đặt x/2=y/5=k

=>x=2k; y=5k

Ta có: xy=10

nên k²=1

Trường hợp 1: k=1

=>x=2; y=5

Trường hợp 2: k=-1

=>x=-2; y=-5

a) \(x>2x\)

\(\Rightarrow x-2x>0\)

\(x\left(1-2\right)>0\)

\(-x>0\)

\(\Rightarrow x< 0\)

b) \(\left(x-1\right)\left(x-2\right)>0\)

\(\Rightarrow\orbr{\begin{cases}x-1>0;x-2>0\\x-1< 0;x-2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>2\\x< 1\end{cases}}\)

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

Mà \(\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(x+1\right)\left(x-4\right)< 0\)

Mà \(x+1>x-4\)

\(\Rightarrow\hept{\begin{cases}x+1>0\\x-4< 0\end{cases}}\)

\(\Rightarrow-1< x< 4\)

d) \(x^3< x^2\)

\(\Rightarrow x^3-x^2< 0\)

\(\Rightarrow x^2\left(x-1\right)< 0\)

\(x^2;x-1\)phải \(\ne\)0

Có  \(x^2>0\); do đó \(x-1< 0\)

\(\Rightarrow x< 1\)