Ôn tập toán 8

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

a: \(=\dfrac{6}{x+1}+\dfrac{4}{x-1}-\dfrac{10}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{6x-6+4x+4-10}{\left(x-1\right)\left(x+1\right)}=\dfrac{10x-12}{\left(x-1\right)\left(x+1\right)}\)

b: \(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{6}{x}=\dfrac{1}{x}+\dfrac{6}{x}=\dfrac{7}{x}\)

\(x^3+4x^2+x-6\)

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

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

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

\(=\left(x-1\right)\left[\left(x^2+2x\right)+\left(3x+6\right)\right]\)

\(=\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

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

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

\(y\left(y-1\right)\left(y-2\right)-y^2+1=0\)

\(\Leftrightarrow y\left(y-1\right)\left(y-2\right)-\left(y-1\right)\left(y+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[y\left(y-2\right)-\left(y+1\right)\right]=0\)

\(\Leftrightarrow\left(y-1\right)\left(y^2-2y-y-1\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y^2-3y-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}y-1=0\\y^2-3y-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}y=1\\y=-0,3027\end{array}\right.\)

\(y\left(y-1\right)\left(y-2\right)-y^2+1=0\)

\(\Leftrightarrow y\left(y-1\right)\left(y-2\right)-\left(y-1\right)\left(y+1\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y^2-2y-y-1\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y^2-3y-1\right)=0\)

Tới đây bạn tự giải :)

Phần giải pt bậc 2 của bạn kết quả đó chỉ là "xấp xỉ" chứ không thể "=" . Tuy nhiên bạn thiếu một nghiệm.

a )\(\left[\begin{array}{nghiempt}x+1\ne0\\2x-3\ne0\end{array}\right.\)

\(ĐKXĐ:x\ne-1,x\ne\frac{3}{2}\)

b ) \(A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{x\left(2x-3\right)}{\left(x+1\right)\left(2x-3\right)}=\frac{x}{x+1}\)

Để \(A=3\) thì :

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

Chúc bạn học tốt

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

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

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

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

Vì \(\left(x-1\right)^2+2\ge2\) với mọi x

\(\Rightarrow-\left[\left(x-1\right)^2+2\right]\le0\)

hay \(-x^2+2x-3\) luôn âm 

Ta có: -x^2+2x-3

            = - (x^2-2x+3)

            = (x^2 - 2x.1+1+2)

            = -[(x-1)^2 +2]

            Do (x-1)^2 +2 > hoặc = 2 với mọi x

           => -[(x-1)^2+2]< hoặc = 0

           =>-x^2+2x-3 luôn âm

Chúc học tốt !

Đặt tổng trên là A có:

     A =  2.4.6.8 +..+ 100.102.104.106 
10A = 2.4.6.8.(10- 0) + 4.6.8.10.(12-2) +....+100.102.104.106.(108-98) 
10A= 2.4.6.8.10 + 4.6.8.10.12 -2.4.6.8.10 +....+ 100.102.104.106.108 -98.100.102.104.106 
10A= 100.102.104.106.108 

\(x^3+4x^2-7x-10\)

\(=x^3-2x^2+6x^2-12x+5x-10\)

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

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

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

\(=\left(x-2\right)\left[x\left(x+1\right)+5\left(x+1\right)\right]\)

\(=\left(x-2\right)\left(x+1\right)\left(x+5\right)\)

x³+4x²-7x-10

= x³-2x²+6x²-12x+5x-10

= x²(x-2)+6x(x-2)+5(x-2)

= (x²+6x+5)(x-2)

= (x²+x+5x+5)(x-2)

= [x(x+1)+5(x+1)](x-2)

= (x+5)(x+1)(x-2)

\(A=2x-x^2\)

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

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

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

Vì\(-\left(x-1\right)^2\)\(\le0\) với mọi x

\(\Rightarrow\)  \(-\left(x-1\right)^2+1\le1\)

Dấu " = " xảy ra khi \(-\left(x-1\right)^2=0\).

                                   \(\Leftrightarrow x=1\)

Vậy \(A_{MAX}=1\) khi \(x=1\)

vậy GTLN của A=1 khi x= 1

Để phân thức \(A=\frac{x^2+5x+4}{x^2+x-12}\) không xác định thì \(x^2+x-12=0\)

\(\Rightarrow x^2+2.\frac{1}{2}x+\frac{1}{4}-12,25=0\)

\(\left(x+\frac{1}{2}\right)^2=12,25\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{7}{2}\\x+\frac{1}{2}=-\frac{7}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-4\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=3\\x=-4\end{array}\right.\)