Ôn tập toán 8

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

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

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

\(=\left(x^2\right)^2+2x^2\cdot x+x^2=\left(x^2+x\right)^2\)

x⁴+2x³+x²

=x²(x²+2x+1)

=x²(x+1)²

Xét vế phải : \(\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{\left(x-2\right)^2}=\frac{a\left(x-2\right)^2}{\left(x+1\right)\left(x-2\right)^2}+\frac{b\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}+\frac{c\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}\)

\(=\frac{a\left(x^2-4x+4\right)+b\left(x^2-x-2\right)+c\left(x+1\right)}{\left(x+1\right)\left(x-2\right)^2}\)

\(=\frac{x^2\left(a+b\right)+x\left(-4a-b+c\right)+\left(4a-2b+c\right)}{\left(x+1\right)\left(x-2\right)^2}\)

So sánh với vế trái, suy ra : 

\(\begin{cases}a+b=2\\-4a-b+c=-1\\4a-2b+c=1\end{cases}\). Giải ra được \(\left(a,b,c\right)=\left(\frac{4}{9};\frac{14}{9};\frac{7}{3}\right)\)

\(x^4+x^3+2x^2+1\)

\(=\left(x^4+2x^2+1\right)+x^3\)

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

 \(x^4+x^3+2x^2+1\)

\(=\left(x^4+2x^2+1\right)+x^3\)

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

a) Đặt \(A=\frac{3n+1}{5n+2}\). Gọi ƯCLN(3n+1 , 5n+2) = d \(\left(d\ge1\right)\) 

Khi đó : \(3n+1⋮d\) và \(5n+2⋮d\)

\(\Rightarrow5\left(3n+1\right)⋮d\) và \(3\left(5n+2\right)⋮d\)

\(\Rightarrow3\left(5n+2\right)-5\left(3n+1\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d\le1\) mà \(d\ge1\Rightarrow d=1\)

Suy ra ƯCLN(3n+1 , 5n+2) = 1 , vậy A là phân số tối giản.

b)  Đặt \(B=\frac{n^3+2n}{n^4+3n^2+1}\) . Gọi ƯCLN(n³+2n , n⁴+3n²+1) = d \(\left(d\ge1\right)\)

Khi đó : \(B=\frac{n\left(n^2+2\right)}{n^2\left(n+2\right)+n^2+1}\)

Ta có : \(n\left(n^2+2\right)⋮d\) và \(n^2\left(n+2\right)+n^2+1⋮d\)

Từ  \(n\left(n^2+2\right)⋮d\) \(\Rightarrow\left[\begin{array}{nghiempt}n⋮d\\n^2+2⋮d\end{array}\right.\)

TH1. Nếu \(n⋮d\) thì ta viết dưới mẫu thức B dưới dạng : 

\(n\left(n^3+3n\right)+1⋮d\) . mà n(n³+3n)\(⋮\)d => \(1⋮d\) \(\Rightarrow d\le1\)

Mà \(d\ge1\Rightarrow d=1\). Lập luận tương tự câu a) , suy ra đpcm

TH2. Nếu \(n^2+2⋮d\) thì ta viết mẫu thức B dưới dạng : 

\(\left(n^4+2n^2\right)+\left(n^2+2\right)-1=\left(n^2+2\right)\left(n^2+1\right)-1⋮d\)

mà  n²+2 \(⋮\)d nên \(1⋮d\Rightarrow d\le1\) mà \(d\ge1\) => d = 1

Lập luận tương tự...

a)Gọi UCLN(3n+1;5n+2) là d

Ta có:

[3(5n+2)]-[5(3n+1)] chia hết d

=>[15n+6]-[15n+5] chia hết d

=>1 chia hết d.Suy ra 3n+1 và 3n+5 là số nguyên tố cùng nhau

=>Phân số tối giản 

b)Gọi d là UCLN(n³+2n;n⁴+3n²+1)

Ta có:

n³+2n chia hết d =>n(n³+2n) chia hết d

=>n⁴+2n² chia hết d (1)

n⁴+3n²-(n⁴+2n²)=n²+1 chia hết d

=>(n²+1)²=n⁴+2n²+1 chia hết d (2)

Từ (1) và (2) => (n⁴+3n²+1)-(n⁴-2n²) chia hết d

=>1 chia hết d

=>d=1.Suy ra n³+2n và n⁴+3n²+1 là 2 số nguyên tố cùng nhau

=>Phân số trên tối giản 

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

\(\Leftrightarrow\left(2x^2+4x+2\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow2.\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow2.\left(x+1\right)^2+\left(y-1\right)^2=0\)

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

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

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

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(4y^2-4y+1\right)=0\)

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

\(\Leftrightarrow\begin{cases}x-2=0\\2y-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=2\\y=\frac{1}{2}\end{cases}\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)\left(4y^2-4y+1\right)=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\2y=1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\y=\frac{1}{2}\end{array}\right.\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)

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

Mà \(\left(x-2\right)^2\ge0,\left(2y-1\right)^2\ge0\)

Suy ra pt trên tương đương với \(\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}\) 

\(\Leftrightarrow\begin{cases}x=2\\y=\frac{1}{2}\end{cases}\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x\left(x^2-9\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow9x=18\)

\(\Leftrightarrow x=2\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+2^3-x.\left(x^2-3^2\right)=26\)

\(\Leftrightarrow x^3+2^3-x^3+9x^2=26\)

\(\Leftrightarrow8+9x=26\)

\(\Leftrightarrow9x=26-8\)

\(\Leftrightarrow9x=18\)

\(\Leftrightarrow x=18:9\)

\(\Leftrightarrow x=2\)

Ta có: 63²+13²-26.13=63²+13²-2.13²=63²-13²=(63-13)(63+13)=50.76

5) \(63^2+13^2-26.13\)

\(=63^2+13^2-2.13.13\)

\(=63^2-13^2\)

\(=\left(63+13\right).\left(63-13\right)=3800\)

\(\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)

\(=\left[x.\left(2x^2-3x+4\right)+2.\left(2x^2-3x+4\right)\right]-\left[x.\left(2x+1\right)-1.\left(2x+1\right)\right]\)

\(=\left(2x^3-3x^2+4x+4x^2-6x+8\right)-\left(2x^3+x-2x-1\right)\)

\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x+2x+1\)

\(=9\)

    \(\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)

 \(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)

\(=9\)

Vậy biểu thức trên không phụ thuộc vào biến x.

a) \(8x^2+30x+7=0\)

\(\Leftrightarrow8\left(x^2+\frac{15}{4}x+7\right)=0\)

\(\Leftrightarrow x^2+\frac{1}{4}x+\frac{7}{2}x+\frac{7}{8}=0\) 

\(\Leftrightarrow x\left(x+\frac{1}{4}\right)+\frac{7}{2}\left(x+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{7}{2}\end{array}\right.\)

b)\(x^3-11x^2+30x=0\)

\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

\(\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\\x-6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

a) \(\Delta=\left(30\right)^2-4.8.7=676>0\) ( PTC2NPB )

\(X_1=\frac{-30+\sqrt{676}}{16}\)

\(X_2=\frac{-30-\sqrt{676}}{16}\)