Vì đồ thị hàm số $y=ax+b$ đi qua điểm $A\left(-1;2\right)$ nên ta có:

   $2=-1.a+b$ suy ra $-a+b=2$

Vi đồ thị hàm số $y=ax+b$ đi qua điểm $B\left(1;4\right)$ nên ta có:

   $4=1.a+b$ suy ra $a+b=4\left(2\right)$

Từ (1) và (2) ta tìm được $a=1;b=3$

Vậy hàm số cần tìm là $y=x+3$.

a)Thay x=2(TMDK) vào bt Q :

Q=2+122−9=−35Q=22−92+1​=−53​

b) P=2x2−1x2+x−x−1x+3x+1=2x2−1x(x+1)−x−1x+3x+1=2x2−1−(x−1)(x+1)+3xx(x+1)=2x2−1−(x2−1)+3xx(x+1)=x2+3xx(x+1)=x(x+3)x(x+1)=x+3x+1P=x2+x2x2−1​−xx−1​+x+13​=x(x+1)2x2−1​−xx−1​+x+13​=x(x+1)2x2−1−(x−1)(x+1)+3x​=x(x+1)2x2−1−(x2−1)+3x​=x(x+1)x2+3x​=x(x+1)x(x+3)​=x+1x+3​

c) M=P.Q=x+3x+1.x+1x2−9=x+3(x−3)(x+3)=1x−3M=P.Q=x+1x+3​.x2−9x+1​=(x−3)(x+3)x+3​=x−31​

M=−12=>1x−3=−12=>x−3=−2=>x=1(TMDK)M=−21​=>x−31​=−21​=>x−3=−2=>x=1(TMDK)

a) $5\left(x+2y\right)-15x\left(x+2y\right)=\left(x+2y\right)\left(5-15x\right)=5\left(x+2y\right)\left(1-3x\right)$

b) 4x2−12x+9=(2x)2−2.2x.3+32=(2x−3)24x2−12x+9=(2x)2−2.2x.3+32=(2x−3)2

c) (3x−2)3−3(x−4)(x+4)+(x−3)3−(x+1)(x2−x+1)=27x3−54x2+18x−8−3(x2−16)+x3−9x2+27x−27−(x3+1)=27x3−54x2+18x−8−3x2+48+x3−9x2+27x−27−x3−1=27x3−57x2+36x+12=3(3x3−19x2+12x+4)(3x−2)3−3(x−4)(x+4)+(x−3)3−(x+1)(x2−x+1)=27x3−54x2+18x−8−3(x2−16)+x3−9x2+27x−27−(x3+1)=27x3−54x2+18x−8−3x2+48+x3−9x2+27x−27−x3−1=27x3−57x2+36x+12=3(3x3−19x2+12x+4)