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a)
\((6x+5)^2(3x+2)(x+1)-35\)
\(=(36x^2+60x+25)(3x^2+5x+2)-35\)
\(=[12(3x^2+5x+2)+1](3x^2+5x+2)-35\)
\(=(12a+1)a-35=12a^2+a-35\) (đặt \(3x^2+5x+2=a)\)
\(=4a(3a-5)+7(3a-5)=(4a+7)(3a-5)\)
\(=(12x^2+20x+15)(9x^2+15x+1)\)
b)
\(8(4x+1)(2x-3)(4x-3)(x+1)-130\)
\(=8[(4x+1)(4x-3)][(2x-3)(x+1)]-130\)
\(=8(16x^2-8x-3)(2x^2-x-3)-130\)
\(=8(8a+21)a-130\) (Đặt \(2x^2-x-3=a\) )
\(=64a^2+168a-130=2(8a-5)(4a+13)\)
\(=2(8x^2-4x+1)(16x^2-8x-29)\)
c)
\((4x+1)(12x-1)(3x+2)(x+1)-4\)
\(=[(4x+1)(3x+2)][(12x-1)(x+1)]-4\)
\(=(12x^2+11x+2)(12x^2+11x-1)-4\)
\(=(a+2)(a-1)-4\) (đặt \(a=12x^2+11x\) )
\(=a^2+a-6=(a-2)(a+3)\)
\(=(12x^2+11x-2)(12x^2+11x+3)\)
d)
\((x+2)(x+3)^2(x+4)-12\)
\(=[(x+2)(x+4)](x+3)^2-12\)
\(=(x^2+6x+8)(x^2+6x+9)-12\)
\(=a(a+1)-12\) (Đặt \(x^2+6x+8=a\) )
\(=a^2+a-12=(a-3)(a+4)=(x^2+6x+5)(x^2+6x+12)\)
\(=(x+1)(x+5)(x^2+6x+12)\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
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
