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\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) \(ĐKXĐ:x\ne\pm3;x\ne-6\)
Với \(x\ne\pm3;x\ne-6\), ta có:
\(P=\left(\dfrac{x}{x-3}-\dfrac{2}{x+3}+\dfrac{x^2}{9-x^2}\right):\dfrac{x+6}{3x+9}\\ =\left(\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2}{\left(x+3\right)\left(x-3\right)}\right)\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x^2+3x-2x+6-x^2}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{3}{x-3}\)
Vậy \(P=\dfrac{3}{x-3}\) với \(x\ne\pm3;x\ne-6\)
b) Ta có: \(2x-\left|4-x\right|=5\)
+) Nếu \(x\le4\Leftrightarrow2x-\left(4-x\right)=5\)
\(\Leftrightarrow2x-4+x=5\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\left(Tm\right)\)
+) Nếu \(x>4\Leftrightarrow2x-\left(x-4\right)=5\)
\(\Leftrightarrow2x-x+4=5\\ \Leftrightarrow x=1\left(Ktm\right)\)
Với \(x\ne\pm3;x\ne-6\)
Khi \(x=3\left(Ktm\right)\rightarrow\text{loại}\)
Vậy khi \(2x-\left|4-x\right|=5\) không có giá trị.
c) Với \(x\ne\pm3;x\ne-6\)
Để P nhận giá trị nguyên
thì \(\Rightarrow\dfrac{3}{x-3}\in Z\)
\(\Rightarrow3⋮x-3\\ \Rightarrow x-3\inƯ_{\left(3\right)}\)
Mà \(Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
Lập bảng giá trị:
| \(x-3\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(0\left(TM\right)\) | \(2\left(TM\right)\) | \(4\left(TM\right)\) | \(6\left(KTM\right)\) |
Vậy để P nhận giá trị nguyên
thì \(x\in\left\{0;2;4\right\}\)
d) Với \(x\ne\pm3;x\ne-6\)
Ta có : \(P^2-P+1=\dfrac{9}{\left(x-3\right)^2}-\dfrac{3}{x-3}+1\)
Đặt \(\dfrac{3}{x-3}=y\)
\(\Rightarrow P^2-P+1=y^2-y+1\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)
\(\Rightarrow P^2-P+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)
Dấu "=" xảy ra khi:
\(\left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow x-3=6\\ \Leftrightarrow x=9\left(TM\right)\)
Vậy \(GTNN\) của biểu thức là \(\dfrac{3}{4}\) khi \(x=9\)
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
hay x<=4
b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)
=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)
=>12x+2+3x+9>=30x+18+48-20x
=>15x+11>=10x+66
=>5x>=55
hay x>=11
Lời giải:
ĐKXĐ: \(x\neq \pm 3\)
a)
Ta có \(P=\frac{2x(3-x)}{(x+3)(3-x)}-\frac{x(x+3)}{(x+3)(3-x)}+\frac{6x}{(3-x)(3+x)}\)
\(=\frac{2x(3-x)-x(x+3)+6x}{(3-x)(3+x)}=\frac{9x-3x^2}{(3-x)(3+x)}=\frac{3x(3-x)}{(3-x)(3+x)}=\frac{3x}{3+x}\)
b) Tại $x=\frac{3}{4}$
\(\Rightarrow P=\frac{3.\frac{3}{4}}{3+\frac{3}{4}}=\frac{3}{5}\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
B xác định \(\Leftrightarrow\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow}x\ne\pm3\)
Vậy B xác định \(\Leftrightarrow x\ne\pm3\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5x-15+3x+9-5x-3}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3}{x+3}\)
a: ĐKXĐ: x<>3; x<>-3; \(x\ne-5\pm\sqrt{34}\)
b: \(=\dfrac{x^2+5x+6+5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2x\left(x-3\right)\left(x+3\right)}{x^2+10x-9}\)
=2x
c: Khi x=1/2 thì A=2*1/2=1
a:
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(Q=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)
\(=\dfrac{3\left(x-3\right)+x+3+18}{x^2-9}\)
\(=\dfrac{3x-9+x+21}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)
b: \(R=Q\cdot x=\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=4+\dfrac{12}{x-3}\)
Để R nguyên thì \(12⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
=>\(x\in\left\{4;2;5;1;6;0;7;-1;9;-3;15;-9\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;1;6;0;7;-1;9;15;-9\right\}\)