Cho hai biểu thức A = \(\dfrac{x^2+x}{3\left(x+3\right)}\) và B = \(\dfrac{1}{x+1}-\dfrac{1}{1-x}-\dfrac{3-x}{x^2-1}\) với x ≠ -3; -1, 1
a) Tính giá trị của biểu thức A khi | x + 4 | = 1
b) Rút gọn biểu thức B
c) Tìm các giá trị của x để B.A <1
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Câu 1:
a) ĐKXĐ: \(x>0;x\ne9\)
Với x=36 (thỏa mãn ĐKXĐ) thì A có giá trị :
\(A=\dfrac{\sqrt{36}+2}{1+\sqrt{36}}=\dfrac{6+2}{1+6}=\dfrac{8}{7}\)
b) Ta có:
\(B=\left(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
c) Ta có:
\(P=A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)
Vì x là số nguyên lớn hơn 0 nên
\(x\ge1\Rightarrow\sqrt{x}\ge1\Rightarrow\sqrt{x}+1\ge2>0\Rightarrow P\le1+\dfrac{3}{2}=\dfrac{5}{2}\)
Dấu bằng xảy ra khi x=1;
Gọi số sản phẩm dự định của xí nghiệp A và B lần lượt là x,y \(\left(x,y\in N;0< x,y< 720\right)\)
Vì tổng sản phẩm dự định là 720 nên ta có phương trình: \(x+y=720\left(1\right)\)
Vì thực tế , xí nghiệp A hoàn thành vượt mức 12% nên số sản phẩm xí nghiệp A thực tế là : \(112\%x=\dfrac{28}{25}x\)
Xí nghiệp B hoàn thành vượt mức 10% nên số sản phẩm xí nghiệp B thực tế là : \(110\%y=\dfrac{11}{10}y\)
Vì tổng số sản phẩm thực tế là 800 nên ta có phương trình: \(\dfrac{28}{25}x+\dfrac{11}{10}y=800\Leftrightarrow56x+55y=40000\left(2\right)\)
Từ (1)(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+y=720\\56x+55y=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=720\\55\cdot720+x=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=400\\y=320\end{matrix}\right.\left(t.m\right)\)
Vậy số sản phẩm 2 xí nghiệp làm theo kế hoạch lần lượt là 400 và 320 sản phẩm
a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)
\(=\dfrac{x^2+1}{x^2+x+1}\)
b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)
=>\(x^2+x+1-x⋮x^2+x+1\)
=>\(x⋮x^2+x+1\)
=>\(x^2+x⋮x^2+x+1\)
=>\(x^2+x+1-1⋮x^2+x+1\)
=>\(-1⋮x^2+x+1\)
=>\(x^2+x+1\in\left\{1;-1\right\}\)
=>\(x^2+x+1=1\)
=>x2+x=0
=>x(x+1)=0
=>\(x\in\left\{0;-1\right\}\)
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
a)Để biểu thức vô nghĩa thì \(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\Leftrightarrow x\notin\left\{-2;1\right\}\)
b) Ta có: \(\dfrac{5x-2}{12}-\dfrac{2x^2+1}{8}=\dfrac{x-3}{6}+\dfrac{1-x^2}{4}\)
\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{24}-\dfrac{3\left(2x^2+1\right)}{24}=\dfrac{4\left(x-3\right)}{24}+\dfrac{6\left(1-x^2\right)}{24}\)
\(\Leftrightarrow10x-4-6x^2-3=4x-12+6-6x^2\)
\(\Leftrightarrow-6x^2+10x-7+6x^2-4x+6=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
Vậy: \(S=\left\{\dfrac{1}{6}\right\}\)
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
1: Ta có: \(A=\left(\dfrac{x^2-16}{x-4}-1\right):\left(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}+\dfrac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+4-1\right):\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}+\dfrac{-x^2+x+2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+3\right):\dfrac{x^2+x-2x-2+x^2-9-x^2+x+2}{\left(x-3\right)\left(x+1\right)}\)
\(=\left(x+3\right):\dfrac{x^2-9}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-3\right)\left(x+1\right)}{x^2-9}\)
\(=x+1\)
ĐKXĐ: \(x\notin\left\{4;3;-1\right\}\)
2: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì \(x+1⋮x^2+x+1\)
\(\Leftrightarrow x^2+x⋮x^2+x+1\)
\(\Leftrightarrow x^2+x+1-1⋮x^2+x+1\)
mà \(x^2+x+1⋮x^2+x+1\)
nên \(-1⋮x^2+x+1\)
\(\Leftrightarrow x^2+x+1\inƯ\left(-1\right)\)
\(\Leftrightarrow x^2+x+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow x^2+x\in\left\{0;-2\right\}\)
\(\Leftrightarrow x^2+x=0\)(Vì \(x^2+x>-2\forall x\))
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Vậy: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì x=0
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
4) Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
5) Ta có: \(B=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
a: Ta có: |x+4|=1
=>x+4=1 hoặc x+4=-1
=>x=-3(loại) hoặc x=-5
Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)
b: \(B=\dfrac{x-1+x+1-3+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)