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B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
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 \(x=9\)
Bài 1:
a) \(x\ne2\)
Bài 2:
a) \(x\ne0;x\ne5\)
b) \(\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
c) Để phân thức có giá trị nguyên thì \(\dfrac{x-5}{x}\) phải có giá trị nguyên.
=> \(x=-5\)
Bài 3:
a) \(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right)\cdot\left(\dfrac{4x^2-4}{5}\right)\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{2\left(2x^2-2\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\cdot2\left(x^2-1\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x^2+3x-x-3\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\left(x+1\right)^2+6-\left(x^2+2x-3\right)\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+6-x^2-2x+3\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+9-x^2-2x\right]\cdot\dfrac{2}{5}\)
\(=\dfrac{2\left(x+1\right)^2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2\left(x^2+2x+1\right)}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2+18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+20}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
c) tự làm, đkxđ: \(x\ne1;x\ne-1\)
ô hô ngộ quá nhìu người bt toán lớp 8 trong khi lớp 7 với lại óc nguyow trở lại r kaka
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
a:\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}=\dfrac{x+1}{x-1}\)
b: Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-1\right)=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
a)P=x2-x+1 đkxđ:x\(\ne\)0;1
b)P=x2-x+1=(x-\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\)\(\ge\)\(\dfrac{3}{4}\) xảy ra dấu = khi x=\(\dfrac{-1}{2}\)
c)Q=\(\dfrac{2x}{P}\)=\(\dfrac{2}{x-1+\dfrac{1}{x}}\)\(\in\)Z đkxđ:x\(\ne\)0
\(\Rightarrow\)2\(⋮\)x-1+\(\dfrac{1}{x}\)\(\Rightarrow\)x-1+\(\dfrac{1}{x}\)\(\in\)U(2)={-2;-1;1;2}
giải ra x\(\in\){-\(\sqrt{\dfrac{5}{4}}\)+\(\dfrac{3}{2}\);\(\sqrt{\dfrac{5}{4}}\)+\(\dfrac{3}{2}\)}
\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
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 ) \(A=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)
\(A=3x+15+x-3-2\left(x+3\right)\)
\(A=4x+10-2x-6\)
\(A=2x+4\)
b ) Để \(A=\dfrac{1}{2}\) thì \(2x+4=\dfrac{1}{2}\), ta có :
\(2x+4=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\dfrac{1}{2}-4\)
\(\Leftrightarrow2x=\dfrac{-7}{2}\)
\(\Leftrightarrow x=\dfrac{-7}{4}\)
Vậy để \(A=\dfrac{1}{2}\) thì \(x=\dfrac{-7}{4}\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài