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\(a,\)\(đkxđ\Leftrightarrow\)\(\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}}\)\(\Rightarrow x\ne\pm3\)
\(b,\)\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(=\frac{5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5x-15+3x+9-5x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x+3}\)
\(c,\)Tại x = 6, ta có :
\(B=\frac{3}{x+3}=\frac{3}{6+3}=\frac{3}{9}=\frac{1}{3}\)
Vậy tại x = 6 thì B = 3
\(d,\)Để \(B\in Z\Rightarrow\frac{3}{x+3}\in Z\Rightarrow x+3\inƯ_3\)
Mà \(Ư_3=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\)TH1 : \(x+3=1\Rightarrow x=-2\)
Th2: \(x+3=-1\Rightarrow x=-4\)
Th3 : \(x+3=3\Rightarrow x=0\)
TH4 \(x+3=-3\Rightarrow x=-6\)
Vậy để \(B\in Z\)thì \(x\in\left\{-6;-4;-2;0\right\}\)
a)Để B đc xác định thì :x+3 khác 0
x-3 khác 0
x^2-9 khác 0
=>x khác -3
x khác 3
b) Kết Qủa BT B là:3/x+3
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
\(ĐKXĐ:x\ne3;x\ne-3;x\ne0\)
b
\(A=\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left[\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right]:\left[\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]\)
\(=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\frac{3x\left(x+3\right)}{-\left(9-3x+x^2\right)}=\frac{-3}{x-3}\)
c
Với \(x=4\Rightarrow A=-3\)
d
Để A nguyên thì \(\frac{3}{x-3}\) nguyên
\(\Rightarrow3⋮x-3\)
Làm nốt.
Đây là 1 bài trong 1 đề t làm nộp gửi thầy nên t đưa ảnh nha,tại lúc đó đề sai nên trong bài giải có vài chữ ko liên quan
Làm tiếp \(M\ge-3\)
\(\frac{x+1}{2x}\ge-3\)
\(\frac{1}{2}+\frac{1}{2x}\ge-3\)
Đến đây dễ r
b, P=x+2x+3−5x2+3x−2x−6+12−xP=x+2x+3−5x2+3x−2x−6+12−x
=x+2x+3−5(x+3)(x−2)−1x−2=x+2x+3−5(x+3)(x−2)−1x−2
=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)
=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)
=x2−4x+3x−12(x+3)(x−2)=x2−4x+3x−12(x+3)(x−2)
=(x−4)(x+3)(x+3)(x−2)=x−4x−2=(x−4)(x+3)(x+3)(x−2)=x−4x−2
c, Để P=−34P=−34
⇔x−4x−2=−34⇔x−4x−2=−34
⇔4(x−4)=−3(x−2)⇔4(x−4)=−3(x−2)
⇔4x−16+3x−6=0⇔4x−16+3x−6=0
⇔7x−22=0⇔7x−22=0
⇔x=227⇔x=227
d, Để P có giá trị nguyên
⇔x−4⋮x−2⇔x−4⋮x−2
⇔(x−2)−2⋮x−2⇔(x−2)−2⋮x−2
⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}
| x−2x−2 | 1 | -1 | 2 | -2 |
| x | 3 | 1 | 4 | 0 |
e,
x2−9=0x2−9=0
⇒x2=9⇒[x=3x=−3⇒x2=9⇒[x=3x=−3
Với x=3,có :
x−4x−2=3−43−2=−11=−1x−4x−2=3−43−2=−11=−1
Với x=-3,có :
x−4x−2=−3−4−3−2=75x−4x−2=−3−4−3−2=75
\(\text{Giải}\)
\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{4x^2-16}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{\left(x+2\right)\left(2x+4\right)}{\left(2x-4\right)\left(2x+4\right)}-\frac{\left(2-x\right)\left(2x-4\right)}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{2x^2+8x+8}{\left(2x-4\right)\left(2x+4\right)}-\frac{4x^2-8+4x}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\frac{2x^2+8x+8-4x^2+8-4x+32}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)
\(A=\frac{4x-2x^2+48}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)
\(A=\frac{2\left(2x-x^2+24\right)}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{\left(2x-4\right)\left(2x+4\right)\left(x-1\right)}\)
\(=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)\left(x-1\right)}=\frac{2x-x^2+24}{\left(x-2\right)\left(x-1\right)}\)
c, Bạn tự giải hệ pt nhé :)
a/ ĐKXĐ: \(x\ne3;-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{3-x}=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}+\frac{-1}{x-3}=\frac{x+2}{x+3}-\frac{5}{\left(x^2-2x\right)+\left(3-6x\right)}\)
\(+\frac{-1}{x-3}=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{-1}{x-3}\)
Đến đây bạn tự quy đồng nhé! Chúc sớm giải được. Cố lên!!!!!