bạn vừa đăng câu này r mà

a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)

\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)

\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

hay \(x\in\left\{1;-1\right\}\)

b: Ta có: \(x^2-7x+10=0\)

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

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

bạn học sd máy tính tìm nghiệm chưa?

a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)

d) \(x^4+7x^2-18=0\\ \Rightarrow\left(x^4-4\right)+\left(7x^2-14\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2\right)+7\left(x^2-2\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x^2=-9\left(loại\right)\end{matrix}\right.\)

b: 4x^2-20x+25=(x-3)^2

=>(2x-5)^2=(x-3)^2

=>(2x-5)^2-(x-3)^2=0

=>(2x-5-x+3)(2x-5+x-3)=0

=>(3x-8)(x-2)=0

=>x=8/3 hoặc x=2

c: x+x^2-x^3-x^4=0

=>x(x+1)-x^3(x+1)=0

=>(x+1)(x-x^3)=0

=>(x^3-x)(x+1)=0

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

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

=>x^2(2x+3)+(2x+3)=0

=>(2x+3)(x^2+1)=0

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

Rồi, nghiệm để làm gì? 

Bài yêu cầu rút gọn và sắp xếp lại phải không bạn?

\(A\left(x\right)=3x^4+10x^2+9\)

\(B\left(x\right)=x^4-5x^2-8\)

\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)

\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)

\(A=7x^3-10x^2+5x-1\)

\(B=-8x^4+7x^2+3\)

a.

\(x^3-7x+6=0\)

\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)

f.

\(x^4-4x^3+12x-9=0\)

\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)

g.

\(x^5-5x^3+4x=0\)

\(\Leftrightarrow x\left(x^4-5x^2+4\right)=0\)

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

\(\Leftrightarrow x\left[x^2\left(x^2-1\right)-4\left(x^2-1\right)\right]=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\end{matrix}\right.\)

`#3107.101107`

`A(x) = 3x - 9x^2 + 4x + 5x^3 + 7x^2 + 1`

`= (3x + 4x) - (9x^2 - 7x^2) + 5x^3 + 1`

`= 7x - 2x^2 + 5x^3 + 1`

`B(x) = 5x^3 - 3x^2 + 7x + 10`

`A(x) - B(x) = 7x - 2x^2 + 5x^3 + 1 - (5x^3 - 3x^2 + 7x + 10)`

`= 7x - 2x^2 + 5x^3 + 1 - 5x^3 + 3x^2 - 7x - 10`

`= (7x - 7x) + (3x^2 - 2x^2) + (5x^3 - 5x^3) - (10 - 1)`

`= x^2 - 9`

`=> C(x) = x^2 - 9`

`C(x) = 0`

`=> x^2 - 9 = 0`

`=> x^2 = 9 => x^2 = (+-3)^2 => x = +-3`

Vậy, nghiệm của đa thức `C(x)` là `x \in {3; -3}.`

a, 3x2 - 8x2 - 2x+3=0

2x(3-8) - 2x+3=0

2x5 - 2x+3=0

2x5 - 2x=0-3=

2x5 - 2x=-3

2x(5-x)=-3

5-x=-3/2

5-x=1,5

x=5-1,5

x=3,5

3,5 nha bn

chúc bn học tốt

happy new year