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\(ĐKXĐ:x\ne\pm1\)
a) \(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
\(\Leftrightarrow A=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
\(\Leftrightarrow A=x-1+x+1-3\)
\(\Leftrightarrow A=2x-3\)
b) Thay x = 3 vào A, ta được :
\(A=2.3-3=3\)
Thay x = 0 vào A, ta được :
\(A=2.0-3=-3\)
c) Để A = 2
\(\Leftrightarrow2x-3=2\)
\(\Leftrightarrow2x=5\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy để \(A=2\Leftrightarrow x=\frac{5}{2}\)
a) 3x(x - 1) + 7x2(x - 1) = 0
<=> x(x - 1)(3 + 7x) = 0
<=> x = 0
hoặc : x - 1 = 0
hoặc 3 + 7x = 0
<=> x = 0
hoặc x = 1
hoặc x = -3/7
b) x2 - 2018x - 2019 = 0
<=> x2 - 2019x + x - 2019 = 0
<=> x(x - 2019) + (x - 2019) = 0
<=> (x + 1)(x - 2019) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)
c) (x + 3)2 - x(x - 2) = 13
<=> x2 + 6x + 9 - x2 + 2x = 13
<=> 8x = 13 - 9
<=> 8x = 6
<=> x= 6/8 = 3/4
a/\(3x\left(x-1\right)+7x^2\left(x-1\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(3x+7x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)x\left(3+7x\right)=0\)
Th1: x - 1 = 0
=> x = 1
Th2: x= 0
Th3: 3 + 7x = 0
=> x= -3/7
\(\Rightarrow x\in\left\{1;0;-\frac{3}{7}\right\}\)
b/ \(x^2-2018x-2019=0\)
\(\Leftrightarrow x^2+x-2019x-2019=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(2019x+2019\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-2019\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2019\right)\left(x+1\right)=0\)
Th1 : x -2019 = 0
=> x =2019
Th2: x + 1 =0
=> x = -1
\(\Rightarrow x\in\left\{2019;-1\right\}\)
c/ \(\left(x+3\right)^2-x\left(x-2\right)=13\)
\(\Leftrightarrow x^2+6x+9-x^2+2x=13\)
\(\Leftrightarrow8x=4\Rightarrow x=\frac{1}{2}\)
a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)
+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)
\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)
+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)
+) \(x+1\ne0\Leftrightarrow x\ne-1\)
+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)
\(\Leftrightarrow x\ne0;x\ne-2\)
+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)
Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)
a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)
\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)
\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)
b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)
Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy ............................
1 M=\(x^2-4xy+4y^2-2x+4y+10\)
=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)
\(=\left(x-2y\right)\left(x-2y-2\right)+10\)
vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)
nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)
\(\Rightarrow\)A\(\ge13\)
dấu "=" xảy ra khi (x-2y)(x-2y-2)=0
\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)
vậy GTNN của M=10 khi x=0; y=0
x=2;y=1
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
Bị tự tin quá khả năng nhẩm mồm, sai em xin lỗi ...
a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)
\(=5x^3-4x-7\)
\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)
\(=13x^3-x^2+4x-5\)
b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)
c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)
\(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)
\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)
\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)
d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )
1
a) x^2+2x-5 b) x^2+x+7 9 (dư 8)
2
x=2; x = -(3*căn bậc hai(7)*i+1)/2;x = (3*căn bậc hai(7)*i-1)/2;
3
a=2
\(Q\left(x\right)=x^2+2x-3=x^2+3x-x-3=\left(x+3\right)\left(x-1\right)\)
Q(x) có nghiệm\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
Áp dụng định lý Bezout:
\(P\left(x\right)⋮Q\left(x\right)\Leftrightarrow\hept{\begin{cases}P\left(-3\right)=0\\P\left(1\right)=0\end{cases}}\)
+) \(P\left(-3\right)=0\Leftrightarrow\left(-3\right)^4+3.\left(-3\right)^3-\left(-3\right)^2-3a+b=0\)
\(\Leftrightarrow81-81-9-3a+b=0\Leftrightarrow3a-b=-9\)(1)
+) \(P\left(1\right)=0\Leftrightarrow1^4+3.1^3-1^2+a+b=0\)
\(\Leftrightarrow1+3-1+a+b=0\Leftrightarrow a+b=-3\)(2)
Lấy (1) + (2), ta được:\(4a=-12\Leftrightarrow a=-3\)
Lúc đó \(b=-3+3=0\)
Vậy a = -3; b = 0
\(P\left(x\right)=x^4+3x^3-x^2+ax+b\)
\(Q\left(x\right)=x^2+2x-3\)
Để phép tính chia hết thì:
\(\Leftrightarrow\hept{\begin{cases}a+5=0\\b-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=-5\\b=3\end{cases}}}\)
Vậy ............