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Bài 1:
a) x≠2x≠2
Bài 2:
a) x≠0;x≠5x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.
=> x=−5x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
a)Đk:\(2x^2+2x\ne0\Rightarrow2x\left(x+1\right)\ne0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\ne0\\x\ne-1\end{array}\right.\) thì phân thức xác định
b)\(\frac{5x+5}{2x^2+2x}=\frac{5\left(x+1\right)}{2x\left(x+1\right)}=\frac{5}{2x}\). Giá trị phân thức =1
\(\Rightarrow\frac{5}{2x}=1\Rightarrow5=2x\Rightarrow x=\frac{5}{2}\)
1
a . X.(X-2)-Y.(X-2)=(X-Y).(X-2)
b .(X2 +1+2X).(X2 +1-2X)
2
3X2 +2X+X2 +2X+1-4X2 -10X+10X+5=(-12)
4X+6= -12
X=9/2
1. a, x2-2x+2y-xy = x(x-2)+y(2-y) = x(x-2)-y(x-2) = (x-y)(x-2)
b, (x2+1)2-4x2 = (x2+1-2x)(x2+1+2x) = (x-1)2(x+1)2
2. x(3x+2)+(x+1)2-(2x-5)(2x+5) = -12
=> (3x2+2x+x2+2x+1)-(2x)2-52 = -12
=> 3x2+2x+x2+2x+1-4x2-25 = -12
=> 4x-24 = -12 => 4x = 12 => x = 3
\(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(Q=x^3+y^3+3xy\left(x+y\right)-2\left(x^2+y^2+2xy\right)+3\left(x+y\right)+10\)
\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
Thay x + y = 5 vào ta có :
\(Q=5^3-2.5^2+3.5+10\)
\(Q=100\)
a) xác định khi x khác +-1
b)
\(A=\left(\frac{\left(2x+1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{\left(x+1\right)}\)
\(A=\left(\frac{\left(2x^2+3x+1\right)+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x+1}\)
\(A=\frac{x^2+5x+8}{\left(x+1\right)^2}=1+\frac{3\left(x+1\right)+4}{\left(x+1\right)^2}\)
c)
GTNN \(B=\frac{3y+4}{y^2}\ge-\frac{9}{16}\)
GTNN \(A=\frac{7}{16}\)
x11+x4+1
= x11+x10+x9-x10-x9-x8+x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1
= x9(x2+x+1)-x8(x2+x+1)+x6(x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x9-x8+x6-x5+x3-x+1)
Bài 2: \(\left(x^2+x\right)^2+4.\left(x^2+x\right)=12\)
Đặt \(t=x^2+x\) vào bt ,ta được:
\(t^2+4t-12=0\)
\(\Leftrightarrow t^2+6t-2t-12=0\)
\(\Leftrightarrow t.\left(t+6\right)-2.\left(t+6\right)=0\)
\(\Rightarrow\left(t+6\right).\left(t-2\right)=0\)
\(\Rightarrow t=-6\) hoặc \(t=2\)
* \(x^2+x=-6\)
\(\Rightarrow x^2+x+6=0\)
mà \(x^2+x+6=x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge0\)
nên t=-6 thì không tìm được giá trị
* \(x^2+x=2\)
\(\Rightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x.\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right)=0\)
\(\Rightarrow x=-2\) hoặc \(x=1\)
Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)
Bài 1:
\(\left(\frac{1}{2x-1}-\frac{1}{2x+1}\right):\frac{4}{10x-5}\)
\(=\left(\frac{2x+1}{4x^2-1}-\frac{2x-1}{4x^2-1}\right)\cdot\frac{10x-5}{4}\)
\(=\frac{2}{\left(2x-1\right)\left(2x+1\right)}\cdot\frac{5\left(2x-1\right)}{4}\)
\(=\frac{5}{2\left(2x+1\right)}\)
Bài 2:
a)Đk:\(2x^2+2x\ne0\Leftrightarrow2x\left(x+1\right)\ne0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ne0\\x\ne-1\end{array}\right.\)
b)\(A=\frac{5x+5}{2x^2+2x}=\frac{5\left(x+1\right)}{2x\left(x+1\right)}=\frac{5}{2x}\)
Phân thức A=1 nghĩ là \(\frac{5}{2x}=1\Rightarrow5=2x\Rightarrow x=\frac{5}{2}\)