Cho A= 3x-5- |x-4|
a)Rút gọn A khi x>4
b) tìm x nếu a= 2016
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a: Ta có: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b: Để P=2 thì \(3\sqrt{x}=2\sqrt{x}+4\)
hay x=16
a: \(A=\dfrac{x\sqrt{x}+1}{x+2\sqrt{x}+1}\)
ĐKXĐ: x>=0
\(A=\dfrac{x\sqrt{x}+1}{x+2\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\)
Thay x=4 vào A, ta được:
\(A=\dfrac{4-2+1}{2+1}=\dfrac{5-2}{3}=1\)
b: M=A*B
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\left(\dfrac{2x+6\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{2x+6\sqrt{x}+7-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)^2}=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)
Để M>2 thì M-2>0
=>\(\dfrac{\sqrt{x}+6-2\sqrt{x}-2}{\sqrt{x}+1}>0\)
=>\(-\sqrt{x}+4>0\)
=>\(-\sqrt{x}>-4\)
=>\(\sqrt{x}< 4\)
=>0<=x<16
c: Để M là số nguyên thì \(\sqrt{x}+6⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1+5⋮\sqrt{x}+1\)
=>\(5⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;-1;5;-5\right\}\)
=>\(\sqrt{x}\in\left\{0;-2;4;-6\right\}\)
=>\(\sqrt{x}\in\left\{0;4\right\}\)
=>\(x\in\left\{0;16\right\}\)
a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)
\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)
\(=-17x+18\)
a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)
c: Để A=3/4 thì 4x-8=3x+6
=>x=14
d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{3;1;4;0;6;-2\right\}\)
a)A=|3x+1|-x-2
=>3x+1=x+2 hoặc 2+x
=>3x+1=x+2 (vì x+2=2+x)
=>A=2x-1
b)Vì |x|=2 =>x=±2
c)A=5 =>2x-1=5
=>2x=6 =>x=3
d)
\(\Rightarrow A\ge-\frac{5}{3}\).Dấu = <=>x=-1/3
Vậy Amin=-5/3 <=>x=-1/3
a: Khi x>4 thì A+3x-5-x+4=2x-1
b: A=2016
=>3x-5-|x-4|=2016(1)
Trường hợp x>=4
=>2x-1=2016
hay x=2017/2(nhận)
Trường hợp 2: x<4
=>3x-5-(4-x)=2016
=>3x-5-4+x=2016
=>4x-9=2016
hay x=2025/4(loại)
a. Khi x > 4
\(A=3x-5-x-4\)
\(=2x-9\)
b. Ta có A = 2016
\(\Rightarrow3x-5-\left|x-4\right|=2016\)
\(\Leftrightarrow-\left|x-4\right|=2016-3x+5\)
\(\Leftrightarrow\left|x-4\right|=3x-2021\)
TH1: \(\left|x-4\right|\ge0\) khi \(x\ge4\)
\(x-4=3x-2021\)
\(\Leftrightarrow-2x=-2017\Leftrightarrow x=\dfrac{2017}{2}\left(tmđk\right)\)
TH2 : \(\left|x-4\right|< 0\) khi \(x< 4\)
\(x-4=2021-3x\)
\(\Leftrightarrow4x=2025\Leftrightarrow x=\dfrac{2025}{4}\left(ktmđk\right)\)
Vậy : Phương trình có tập nghiệm \(S=\left\{\dfrac{2017}{2}\right\}\)