Các bn giúp mình với mình đang cần gấp

nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên

Bài 1:

16:

=>15-30x-8+8x=4x-9

=>-22x+7=4x-9

=>-26x=-16

=>x=8/13

15: \(\Leftrightarrow12x-6+12-4x=10-6x\)

=>8x+6=10-6x

=>14x=4

=>x=2/7

14: \(\Leftrightarrow12x-18+9-15x=8x-9\)

=>-3x-9=8x-9

=>x=0

13: \(\Leftrightarrow10x+15x-10=25-10x\)

=>25x-10=25-10x

=>35x=35

=>x=1

12: \(\Leftrightarrow15x-18-4x+10=11x-10\)

=>11x-8=11x-10(loại)

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

b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)

c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)

Câu 1:

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

Câu 2:

\(a,\left(3x+ay+b\right)\left(x+cy+d\right)\\ =3x^2+3xcy+3xd+axy+acy^2+ayd+bx+bcy+bd\\ =3x^2+xy\left(3c+a\right)+x\left(b+3d\right)+y\left(ad+bc\right)+acy^2+bd\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}3c+a=-22\\b+3d=-4\end{matrix}\right.\\ad+bc=8\\\left\{{}\begin{matrix}ac=7\\bd=1\end{matrix}\right.\end{matrix}\right.\)

Xét \(bd=1\Leftrightarrow\left[{}\begin{matrix}b=1;d=1\\b=-1;d=-1\end{matrix}\right.\)

Với \(b=1;d=1\Leftrightarrow b+3d=1+3\cdot1=4\left(ktm\right)\)

Với \(b=-1;d=-1\Leftrightarrow b+3d=-1-3=-4\left(tm\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3c+a=-22\\-a-c=8\\ac=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\c=-7\end{matrix}\right.\)

Vậy \(3x^2-22xy-4x+8y+7y^2+1=\left(3x-y-1\right)\left(x-7y-1\right)\)

Cái chỗ ngoặc nhọn mà 5 dòng á a ko thấy trong cái phần công thức nên là ghi z chứ nó có 5 dòng đó nha

câu b tương tự, lười wa 😴

c, x⁴+6x³+11x²+6x+1
=x⁴+6x³+9x²+2x²+6x+1
=x⁴+9x²+1+6x³+2x²+6x
=(x²)²+(3x)²+1²+2.x².3x+2.x².1+2.3x.1       (1)
Áp dụng hằng đẳng thức (a+b+c)²=a²+b²+c²+2ab+2ac+2bc
=> (1)=(x²+3x+1)²

Câu a nhé bạn:
a,   3x²−22xy−4x+8y+7y²+1
=3x²-21xy-xy-3x-x+7y+y+7y²+1
=(3x²−21xy−3x)−(xy-7y²-y)−(x-7y-1)
=3x(x−7y−1)−y(x−7y−1)−(x−7y−1)
=(3x−y−1)(x−7y−1)

Cho mk hs bn Diệu Thúy để lm đc như câu a) thì có pp j k bn? Hay chỉ lag ngồi nhóm lại thôi hả bn? Trả lời nhanh dùm mk nha😘

b.\(x^3+6x^2+11x+6=0\)

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

\(\Leftrightarrow x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\)

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

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

\(\Leftrightarrow\)\(x+1=0\)hoặc \(x+2=0\)hoặc \(x+3=0\)

\(\Leftrightarrow\)...... tự viết nha bn

a)   (6x⁵ - 3x²):3x - (4x² + 8x):4x = 5 

\(\Rightarrow\)2x⁴ - x - x - 2 = 5

\(\Rightarrow\)2(x⁴ - x -1) = 5

\(\Rightarrow\)x⁴ - 2x²  + 1 + 2x² - 2 = 2.5

\(\Rightarrow\)(x² - 1)² + 2(x² - 1)  + 1 - \(\frac{7}{2}\) = 0

\(\Rightarrow\)x⁴ = \(\frac{7}{2}\)

\(\Rightarrow\)x  = \(\pm\)\(\sqrt[4]{\frac{7}{2}}\) 

b)   x³ + 6x² + 11x +6 = 0

\(\Rightarrow\)x³ + 6x² + 12x + 8 - x - 2 = 0

\(\Rightarrow\)(x + 2)³ - (x + 2) = 0

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

\(\Rightarrow\)x + 2 = 0    \(\Rightarrow\)x = -2 

         x - 1 =0        \(\Rightarrow\)x = 1

         x + 3 = 0       \(\Rightarrow\)x = -3

Vay.....