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a, ĐKXĐ :\(x\ne3;x\ne-3\)
b, \(P=\frac{3\cdot\left(x-3\right)}{\left(x-3\right)\cdot\left(x+3\right)}+\frac{x+3}{\left(x+3\right)\cdot\left(x-3\right)}+\frac{18}{\left(x+3\right)\cdot\left(x-3\right)}\)
\(=\frac{3x-9+x+3+18}{\left(x+3\right)\cdot\left(x-3\right)}\)\(=\frac{4x+12}{\left(x-3\right)\cdot\left(x+3\right)}\)
\(=\frac{4\cdot\left(x+3\right)}{\left(x+3\right)\cdot\left(x-3\right)}=\frac{4}{x-3}\)
c, Với P = 4 \(\Rightarrow\frac{4}{x-3}=4\Rightarrow4=4\cdot\left(x-3\right)\)\(\Rightarrow1=x-3\Rightarrow x=4\)
\(x\ne+-3\)
\(3\left(x-3\right)+1\left(x+3\right)+18\)
3x-9+x+3+18
4x+15
x=-15/4
Bạn viết biểu thức A ra đi rồi bọn mình mới làm được chứ -.-
Đk : \(x\ne\pm3\)
Để B>A
\(\Leftrightarrow\frac{3}{x+3}>4\)
Rõ ràng: \(x+3>0\)
\(\Rightarrow\frac{3}{x+3}>4\)
\(\Leftrightarrow3>4\left(x+3\right)\)
\(\Leftrightarrow3>4x+12\)
\(\Leftrightarrow-9>4x\)
\(\Leftrightarrow x< \frac{-9}{4}\)
KL: \(x\in Z,x< \frac{-9}{4},x\ne\pm3\)
a) ĐKXĐ: x \(\ne\pm3\)
b) = \(\frac{3\left(x-3\right)+x+3+18}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4}{x-3}\)
c) P = 4 hay \(\frac{4}{x-3}=4\)=> x - 3 = 1 <=> x = 4 (TM)
Vậy ...
bài1 A=\(\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
=\(\left(-\frac{x-3\cdot\left(x+3\right)^2}{\left(x+3\right)^2\cdot\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
=\(-\frac{x}{x+3}\cdot\frac{x+3}{3x^2}=\frac{-1}{3x}\)
b) thế \(x=-\frac{1}{2}\)vào biểu thức A
\(-\frac{1}{3\cdot\left(-\frac{1}{2}\right)}=\frac{2}{3}\)
c) A=\(-\frac{1}{3x}< 0\)
VÌ (-1) <0 nên 3x>0
x >0
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
A= \(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}=\frac{3x-9}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{18}{\left(x+3\right)\left(x-3\right)}\)
= \(\frac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}=\frac{4}{x-3}\)
b) để A=4 thì \(\frac{4}{x-3}=4\)=> x-3=1=|> x=4
Cây dừa, cây hành, cây ngô