a) \(=(x-y)[7x-(x-y)]=(x-y)(6x+y)\)

b) (sửa đề) \(ax-ay+by-by\)

\(=ax-ay=a(x-y)\)

Mình cho kết quả luôn nha!

a) \(x=3\)

b) \(x=-7,3\)

c) \( x=2,1\)

d) \(x=-0,3\)

Kết quả: \(\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)

a) Thay \(k=0\) vào phương trình ta có:

\(4x^2-25+0^2+4.0.x=0\)

\(⇔4x^2-25=0\)\(⇔(2x)^2-5^2=0\)

\(⇔(2x-5)(2x+5)=0\)

\(⇔\left[\begin{array}{} 2x-5=0\\ 2x+5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{5}{2}\\ x=-\frac{5}{2} \end{array}\right.\)

b) Thay \(k=-3\) vào phương trình ta có:

\(4x^2-25+(-3)^2+4.(-3).x=0\)

\(⇔4x^2-12x-16=0\)

\(⇔4x^2-16x+4x-16=0\)

\(⇔4x(x-4)+4(x-4)=0\)

\(⇔(x-4)(4x+4)=0\)

\(⇔\left[\begin{array}{} x-4=0\\ 4x+4=0 \end{array}\right.⇔\left[\begin{array}{} x=4\\ x=-1 \end{array}\right.\)

c) Thay \(x=-2\) vào pt ta có:

\(4.(-2)^2-25+k^2+4k.(-2)=0\)

\(⇔k^2-8k-9=0\)

\(⇔k^2-9k+k-9=0\)

\(⇔k(k-9)+(k-9)=0\)

\(⇔(k-9)(k+1)=0\)

\(⇔\left[\begin{array}{} k-9=0\\ k+1=0 \end{array}\right.⇔\left[\begin{array}{} k=9\\ k=-1 \end{array}\right.\)

a) \(⇔-4x^2-4x+7x+7=0\)

\(⇔-4x(x+1)+7(x+1)=0\)

\(⇔(x+1)(-4x+7)=0\)

\(⇔\left[\begin{array}{} x+1=0\\ -4x+7=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=\frac{7}{4} \end{array}\right.\)

b) \(⇔4x^2-8x+3x-6=0\)

\(⇔4x(x-2)+3(x-2)=0\)

\(⇔(x-2)(4x+3)=0\)

\(⇔\left[\begin{array}{} x-2=0\\ 4x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=2\\ x=-\frac{3}{4} \end{array}\right.\)

a) \(⇔x^2-9x+20=12 \)

\(⇔x^2-9x+8=0\)

\(⇔x^2-x-8x+8=0\)

\(⇔(x-1)(x-8)=0\)

\(⇔\left[\begin{array}{} x-1=0\\ x-8=0 \end{array}\right.⇔\left[\begin{array}{} x=1\\ x=8 \end{array}\right.\)

b) \(⇔4x^2-12x+8=3\)

\(⇔4x^2-12x+5=0\)

\(⇔(2x-1)(2x-5)=0\)

\(⇔\left[\begin{array}{} 2x-1=0\\ 2x-5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{5}{2} \end{array}\right.\)

c) \(⇔x^2+x-30=42\)

\(⇔x^2+x-72=0\)

\(⇔(x-9)(x+8)=0\)

\(⇔\left[\begin{array}{} x-9=0\\ x+8=0 \end{array}\right.⇔\left[\begin{array}{} x=9\\ x=-8 \end{array}\right.\)

d) \(⇔2x^2+5x-3=-6\)

\(⇔2x^2+5x+3=0\)

\(⇔(x+1)(2x+3)=0\)

\(⇔\left[\begin{array}{} x+1=0\\ 2x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=-\frac{3}{2} \end{array}\right.\)

+) \(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{(2x-1)(2x+1)}{x^3-1}\)(ĐKXĐ: x ≠ 1)

\(⇔\frac{2x^2+2x+2}{(x-1)(x^2+x+1)}+\frac{2x^2+x-3}{(x-1)(x^2+x+1)}=\frac{(2x-1)(2x+1)}{(x-1)(x^2+x+1)}\)

\(⇔\frac{4x^2+3x-1}{(x-1)(x^2+x+1)}=\frac{4x^2-1}{(x-1)(x^2+x+1)}\)

\(⇔4x^2+3x-1=4x^2-1\)

\(⇔4x^2-4x^2+3x=-1+1\)

\(⇔3x=0⇔x=0\)(thỏa mãn ĐKXĐ)

+) \(\frac{x^3-(x-1)^3}{(4x+3)(x-5)}=\frac{7x-1}{4x+3}-\frac{x}{x-5}\)(ĐKXĐ: x ≠ \(\frac{-3}{4}\); x ≠ 5)

\(⇔\frac{3x^2-3x+1}{(4x+3)(x-5)}=\frac{7x^2-36x+5}{(4x+3)(x-5)}-\frac{4x^2+3x}{(4x+3)(x-5)}\)

\(⇔\frac{3x^2-3x+1}{(4x+3)(x-5)}=\frac{3x^2-39x+5}{(4x+3)(x-5)}\)

\(⇔3x^2-3x+1=3x^2-39x+5\)

\(⇔3x^2-3x^2-3x+39x=5-1\)

\(⇔36x=4⇔x=\frac{1}{9}\)(thỏa mãn ĐKXĐ)

\(x^2(x-5)+x^2-4x-5=0\)

\(⇔x^2(x-5)+x^2+x-5x-5=0 \)

\(⇔x^2(x-5)+(x+1)(x-5)=0 \)

\(⇔(x-5)(x^2+x+1)=0\)

\(⇔\left[\begin{array}{} x-5=0\\ x^2+x+1=0(mà:x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}>0) \end{array}\right.\)

\(⇔x=5\)

Vậy pt có 1 nghiệm là x=5