\(âP\left(x\right)=13x^3+4x^2-11x-2\)

\(b.Q\left(x\right)=x^3+9x-5\)
\(c.A\left(x\right)=14x^3-x^2+10x+14\)
\(d.B\left(x\right)=2x^2+x+3\)

Nếu ol thì tham khảo nah nguoiemtinhthong.

1.1

2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)

Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0

a=2ba=2b v a=13ba=13b

Các bạn tự giải quyết tiếp nhé.

1.2

TXĐ D=[1;+∞)D=[1;+∞)

đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0

pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0

⇔a=b⇔a=b v a=23ba=23b

...

1.3

D=[3;+∞)D=[3;+∞)

Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0

pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2

⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0

⇒a=5b⇒a=5b
...

1.4

ĐK

⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)

⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)

Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)

⇔2a2+2b2=3ab

1.5

Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)

⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x

⇔t2−t−4x2+2x=0t2−t−4x2+2x=0

Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2

⇒t=1−2xt=1−2x hoặc t=2xt=2x

1.1

2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1

2(.2+x+1)+3(x-1)

3a+b=11a2-19b2

tóm tắt