ai giải giúp em đi ạ em đang cần gấp lắm ạ 

Giup cai j ? Cau nao ?

Đề số 3.

1.

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

\(=20x^3-8x^2+12x\)

b.\(\left(x-2\right)\left(x^2-3x+5\right)\)

\(=x^3-3x^2+5x-2x^2+6x-10\)

\(=x^3-5x^2+11x-10\)

c,\(\left(10x^4-5x^3+3x^2\right):5x^2\)

\(=2x^2-x+\dfrac{3}{5}\)

d,\(\left(x^2-12xy+36y^2\right):\left(x-6y\right)\)

\(=\left(x-6y\right)^2:\left(x-6y\right)\)

\(=x-6y\)

2.

a,\(x^2+5x+5xy+25y\)

\(=\left(x^2+5x\right)+\left(5xy+25y\right)\)

\(=x\left(x+5\right)+5y\left(x+5\right)\)

\(=\left(x+5y\right)\left(x+5\right)\)

b,\(x^2-y^2+14x+49\)

\(=\left(x^2+14x+49\right)-y^2\)

\(=\left(x+7\right)^2-y^2\)

\(=\left(x+7-y\right)\left(x+7+y\right)\)

c,\(x^2-24x-25\)

\(=x^2+25x-x-25\)

\(=\left(x^2-x\right)+\left(25x-25\right)\)

\(=x\left(x-1\right)+25\left(x-1\right)\)

\(=\left(x+25\right)\left(x-1\right)\)

3.

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{5}\) hoặc \(x=3\)

b.\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(3x^2-15x-\left(2x+3x^2-2-3x\right)=30\)

\(3x^2-15x-2x-3x^2+2+3x=30\)

\(-14x+2=30\)

\(-14x=28\)

\(x=-2\)

c,\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\)

\(x^2+5x+6-x^2-5x+2x+10=0\)

\(2x+16=0\)

\(2x=-16\)

\(x=-8\)

Mình học chật hình không giúp bạn được.Xin lỗi!

23.27. \(x^2-y^2-2x+1\)

\(=\left(x-1\right)^2-y^2\)

\(=\left(x-1-y\right)\left(x-1+y\right)\)

23.25.

\(\left(x^2-4x\right)^2+\left(x-2\right)^2-10\)

\(=\left(x^2-4x\right)^2-4+\left(x-2\right)^2-6\)

\(=\left(x^2-4x+4\right)\left(x^2-4x-4\right)+x^2-4x+4-6\)

\(=\left(x^2-4x+4\right)\left(x^2-4x-10\right)\)

23.23

\(x^3-2x^2-6x+27\)

\(=\left(x^3+27\right)-2x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9-2x\right)\)

\(=\left(x+3\right)\left(x^2-5x+9\right)\)

23.27

\(x^2-y^2-2x+1\)

\(=\left(x^2-2x+1\right)-y^2\)

\(=\left(x-1\right)^2-y^2\)

\(=\left(x-1-y\right)\left(x-1-y\right)\)

18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)

19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)

\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)

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

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

21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)

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

\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)

23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

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

ta có :

1) \(\frac{x-y}{z-y}=-10\Leftrightarrow x-y=10\left(y-z\right)\)

\(\Leftrightarrow x-y=10y-10z\)

\(\Leftrightarrow x=11y-10z\)

Thay x=11y-10z vào biểu thức \(\frac{x-z}{y-z}\), ta có:

\(\frac{11y-10z-z}{y-z}=\frac{11y-11z}{y-z}=\frac{11\left(y-z\right)}{y-z}=11\)

Chá quá, có ghi nhìn không rõ đề

2) \(2x^2=9x-4\)

\(\Leftrightarrow2x^2-9x+4=0\)

\(\Leftrightarrow2x^2-8x-x+4=0\)

\(\Leftrightarrow2x\left(x-4\right)-1\left(x-4\right)\)

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

\(\Leftrightarrow2x-1=0\) hoặc x-4=0

1) 2x-1=0<=>x=1/2

2)x-4=0<=>x=4(Loại)

=> x=1/2

Bài 4: (ý c chuyên toán Hình vào làm nốt nhé @@)

Mình làm 1 bài thôi nhé

Bài 5 

\(a.1-2y+y^2=\left(1-y\right)^2\)

\(b.\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x-4\right)\left(x+6\right)\)

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

\(d.27+27x+9x^2+x^3=3^3+3.3^3.x+3.3.x^2+x^3=\left(3+x\right)^3\)

\(f.8x^3-12x^2y+6xy-y^3=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y-y^3=\left(2x-y\right)^3\)

Bài 4 : 

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

b, bạn xem lại đề nhé 

c, \(x^2-4x+4-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)

d, \(5x+5-x^2+1=5\left(x+1\right)+\left(1-x\right)\left(x+1\right)=\left(x+1\right)\left(6-x\right)\)

Bài 2 :

a ) \(25-20x+4x^2=0\)

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

\(\Leftrightarrow5-2x=0\Rightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

a,\(\left(-2x^2+3x\right)\left(x^2-x+3\right)\\ \Leftrightarrow-2x^4+2x^3-6x^2+3x^3-3x^2+9x\\ \Leftrightarrow-2x^4+5x^3-3x^2+3x\)

\(b,x\left(x-2\right)\left(x+2\right)-\left(x-3\right)\left(x^2+3x+9+6\right)+6\left(x+1\right)^2=15\\ \Leftrightarrow x\left(x^2-4\right)-\left(x^3-27\right)+6\left(x^2+2x+1\right)=15\\ \Leftrightarrow x^3-4x-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow6x^2+8x+18=0\\ \Leftrightarrow6\left(x^2+\dfrac{4}{3}x+3\right)=0\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^2+\dfrac{23}{9}=0\)

Với mọi x thì \(\left(x+\dfrac{2}{3}\right)^2\ge0\Rightarrow\left(x+\dfrac{2}{3}\right)^2+\dfrac{23}{9}>0\)

Do đó ko tìm đc giá trị nào của x thỏa mãn đề bài

Vậy..