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mk chỉ lm đc câu b thôi !
mk k viết đề đâu nha !:
=(x4-8x2+16)+(5x2-20x+20)+(3x2+12x+12)+15
=(x2-4)2+5(x-2)2+3(x-2)2+15
=[(x-2)2+3][(x-2)2+5]
=(x2-4x+7)(x2+4x+9)
đúng 100 %
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này giải hệ số bất định.
Ta có:
\(x^4-8x+63\)
\(=x^4+4x^3-4x^3+9x^2-16x^2+7x^2-36x+28x+63\)
\(=\left(x^4-4x^3+7x^2\right)+\left(4x^3-16x^2+28x\right)+\left(9x^2-36x+63\right)\)
\(=x^2\left(x^2-4x+7\right)+4x\left(x^2-4x+7\right)+9\left(x^2-4x+7\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=\left(x+1\right)^4+x^4+x^2+1+2x^3+2x^2+2x\)
\(=\left(x+1\right)^4+x^4+3x^2+1+2x^3+2x\)
a) \(x^4-7x^3+14x^2-7x+1\)(1)
Giả sử x khác 0, khi đó :
\(\left(1\right)\Leftrightarrow x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(\Leftrightarrow x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+14\right]\)
\(\Leftrightarrow x^2\left[\left(x^2+2\cdot x\cdot\dfrac{1}{x}+\dfrac{2}{x^2}\right)-2-7\left(x+\dfrac{1}{x}\right)+14\right]\)
\(\Leftrightarrow x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
Đặt \(x+\dfrac{1}{x}=a\)
pt \(\Leftrightarrow x^2\left(a^2-7a+12\right)\)
\(\Leftrightarrow x^2\left(a^2-3a-4a+12\right)\)
\(\Leftrightarrow x^2\left[a\left(a-3\right)-4\left(a-3\right)\right]\)
\(\Leftrightarrow x^2\left(a-3\right)\left(a-4\right)\)
\(\Leftrightarrow x^2\left(x+\dfrac{1}{x}-3\right)\left(x+\dfrac{1}{x}-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(x^2-2x-24=0\Leftrightarrow x^2-2x+1-25=0\)
\(\Leftrightarrow\left(x-1\right)^2-5^2=0\Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
b)\(x^2+8x+12=0\Leftrightarrow x^2+8x+16-4=0\)
\(\Leftrightarrow\left(x+4\right)^2-2^2=0\Leftrightarrow\left(x+4-2\right)\left(x+4+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\Leftrightarrow\hept{\begin{cases}x=-2\\x=-6\end{cases}}\)
c)\(4x^2+4x-63=0\Leftrightarrow4x^2+4x+1-64=0\)
\(\Leftrightarrow\left(2x+1\right)^2-8^2=0\Leftrightarrow\left(2x+1-8\right)\left(2x+1+8=0\right)\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+9\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =x^4-4x^3+7x^2+4x^3-16x^2+28x+9x^2-28x+63
=(x^2-4x+7)(x^2+4x+9)
b: =(x^2+2x+1)^2+(x^2+x+1)^2
=x^4+4x^2+1+4x^3+2x^2+4x+x^4+x^2+1+2x^3+2x^2+2x
=2x^4+6x^3+9x^2+6x+2
=(x^2+2x+2)(2x^2+2x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
P(x) = x4 - 8x + 63
P(x) phân tích đươc thành nhân tử có dạng (x2 + ax + b)(x2 + cx + d)
= x4 + (a + c)x3 + (b + ac + d)x2 + (ad + bc)x + bd
Đồng nhất đa thức trên với đa thức đã cho, ta có:
\(\left\{{}\begin{matrix}a+c=0\\b+ac+d=0\\ad+bc=-8\\bd=63\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\\c=4\\d=9\end{matrix}\right.\)
Vậy P(x) = (x2 - 4x + 7)(x2 + 4x + 9)
Q(x) = 2x4 - 7x3 + 17x2 - 20x + 14
Q(x) phân tích đươc thành nhân tử có dạng (x2 + ax + b)(2x2 + cx + d)
= 2x4 + (2a + c)x3 + (2b + ac + d)x2 + (ad + bc)x + bd
Đồng nhất đa thức trên với đa thức đã cho, ta có:
\(\left\{{}\begin{matrix}2a+c=-7\\2b+ac+d=17\\ad+bc=-20\\bd=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=2\\c=-3\\d=7\end{matrix}\right.\)
Vậy Q(x) = (x2 - 2x + 2)(2x2 - 3x + 7)
![](https://rs.olm.vn/images/avt/0.png?1311)
xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai
ĐK : \(X\ne-1;-3;-7;-9\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)
\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)
\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)
\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)
\(2\left(x+1\right)\left(x+9\right)=40\)
\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)
\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)
\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)
\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn )
Vậy ...............
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
\(x^4-8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7x\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7x\right)\)
chúc bạn học tốt ! ^-^