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\(2x^2-10x+8\)
\(=2x^2-8x-2x+8\)
\(=x\left(2x-8\right)-\left(2x-8\right)\)
\(=\left(x-1\right)\left(2x-8\right)\)
\(=2\left(x-1\right)\left(x-4\right)\)
Chúc bạn học tốt.
\(2x^2-10x+8\)
\(=2x^2-2x-8x+8\)
\(=\left(2x^2-2x\right)-\left(8x-8\right)\)
\(=2x\left(x-1\right)-8\left(x-1\right)\)
\(=\left(x-1\right)\left(2x-8\right)\)
\(=2\left(x-1\right)\left(x-4\right)\)
\(\left(a\right)x^4-2x^3+3x^2-2x+1\)
\(\text{phân tích đa thức thành nhân tử:}\)
b) c) (x2 + x)(x2 + x + 1) - 2
d) (x + 1)(x + 2)(x + 3)(x + 4) - 3
a) \(x^3-5x^2+8x-4=\left(x^3-x^2\right)-4\left(x^2-x\right)+4\left(x-1\right)=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\) chia hết cho 2x-3 => 7 chia hết cho 2x -3
=> 2x -3 thuộc U(7) ={-7;-1;1;7}
+2x-3 =-7 => x =-2
+2x-3 =-1 => x =1
+2x-3 =1 => x =2
+2x -3 =7 => x =5
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
minh moi bn vao link nay dang ky roi tra loi minigame nha : https://alfazi.edu.vn/question/5b7768199c9d707fe5722878
a, x4 - 3x3 - x + 3
= (x4 - x) - (3x3 - 3)
= x(x3 - 1) - 3(x3 - 1)
= (x - 3)(x3 - 1)
b, x2 - x - 12
= x2 - x - 16 + 4
= (x2 - 16) - (x - 4)
= (x2 - 42) - (x - 4)
= (x + 4)(x - 4) - (x - 4)
= (x + 4 - 1)(x - 4)
= (x + 3)(x - 4)
c, x2 - 7x + 12
= x2 - 3x - 4x + 12
= (x2 - 3x) - (4x - 12)
= x(x - 3) - 4(x - 3)
= (x - 4)(x - 3)
d, x2 - 2x - 8
= x2 - 4x + 2x - 8
= (x2 - 4x) + (2x - 8)
= x(x - 4) + 2(x - 4)
= (x + 2)(x - 4)
5, x2 - 10x + 21
= x2 - 3x - 7x + 21
= (x2 - 3x) - (7x - 21)
= x(x - 3) - 7(x - 3)
= (x - 7)(x - 3)
f, x7 - x2 - 1
= t không bt
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(x^2-x+1=x^2-2\times x\times\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)\(\frac{3}{4}\)
= \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
vì \(\left(x-\frac{1}{2}\right)^2\ge0\)
=> \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
vậy Min A= \(\frac{3}{4}\)dấu bằng xảy ra khi và chỉ khi \(x=\frac{1}{2}\)
ở trên bạn bỏ hộ mk 1 phân số \(\frac{3}{4}\)đi nhé mk viết thừa.
a, x2 - 10x + 24 = x2 - 2.x.5 + 25 - 1 = (x-5)2 - 1 = (x - 5 - 1).(x - 5 + 1) = (x - 6).(x - 4)
b, x2 - x - 30 = x2 - 6x + 5x - 30 = x.(x - 6) + 5.(x - 6) = (x - 6).(x + 5)
c, x2 + 9x + 8 = x2 + x + 8x + 8 = x.(x + 1) + 8.(x + 1) = (x + 1).(x + 8)
Câu d đề có phải sai không vậy bạn, mình giải theo đề này nha: x2 - 10x + 16 = x2 - 8x - 2x +16 = x.(x - 8) - x.(x - 8) = (x - 2).(x - 8)
a/ x4 +5x3 +10x-4
=(x4- 4)+(5x3 + 10x)
=(x2+2) (x2-2) + 5x(x2 +2 )
=(x2+2)(x2 -2 +5x)
b/x5 - x4 +x3 -x2 +x-1
=x4(x-1)+x3(x-1)+(x-1)
=(x-1)(x4+x3+1)
\(=\dfrac{1}{2}\left(4x^2-20x-2\right)=\dfrac{1}{2}\left[\left(4x^2-20x+25\right)-27\right]\)
\(=\dfrac{1}{2}\left[\left(2x-5\right)^2-\left(3\sqrt[]{3}\right)^2\right]=\dfrac{1}{2}\left(2x-5-3\sqrt{3}\right)\left(2x-5+3\sqrt{3}\right)\)