Biết \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\), \(a+b+c=abc\)
Tính \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
Cho hệ PT: với m là tham số hệ có 2 nghiệm (x;y) .TÌm GTNN của
giúp mình với ạ
1. cmr nếu a,b,c khác 0, a+b+c=0 thì \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=0\)
2. rút gọn \(P=\dfrac{x^4}{1-x}+x^3+x^2+x+1\)
2: \(P=\dfrac{x^4}{1-x}+x^3+x^2+x+1\)
\(=-\dfrac{x^4}{x-1}+\left(x^3+x^2+x+1\right)\)
\(=\dfrac{-x^4+\left(x-1\right)\left(x^3+x^2+x+1\right)}{x-1}\)
\(=\dfrac{-x^4+x^4-1}{x-1}=\dfrac{-1}{x-1}\)
1: \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=\dfrac{0}{abc}=0\)
help mee! pls
ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{-5-\sqrt{17}}{2}\\x>=\dfrac{-5+\sqrt{17}}{2}\end{matrix}\right.\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
=>\(\left(x^2+5x+4\right)-3\sqrt{x^2+5x+2}-6=0\)
=>\(\left(x^2+5x+2\right)-3\sqrt{x^2+5x+2}-4=0\)
=>\(\left(\sqrt{x^2+5x+2}-4\right)\left(\sqrt{x^2+5x+2}+1\right)=0\)
=>\(\sqrt{x^2+5x+2}-4=0\)
=>\(\sqrt{x^2+5x+2}=4\)
=>\(x^2+5x+2=16\)
=>\(x^2+5x-14=0\)
=>(x+7)(x-2)=0
=>\(\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
tìm số hữu tỉ x trong các tỉ lệ thức sau :
2x+4/5 = 2x+1/10 ( / : phần )
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{2x+4}{5}=\dfrac{2x+1}{10}=\dfrac{\left(2x+4\right)-\left(2x+1\right)}{5-10}=-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{2x+4}{5}=-\dfrac{3}{5}\)
\(\Rightarrow2x+4=-3\)
\(\Rightarrow2x=-7\)
\(\Rightarrow x=-\dfrac{7}{2}\)
x, y là các số thỏa mãn \(9x^2+4y^2=20xy;2y< 3x< 0\). Tính \(A=\dfrac{3x^2+xy-2y^2}{3x^2+5xy+2y^2}\)
\(9x^2+4y^2=20xy\)
=>\(9x^2-20xy+4y^2=0\)
=>\(9x^2-18xy-2xy+4y^2=0\)
=>9x(x-2y)-2y(x-2y)=0
=>(x-2y)(9x-2y)=0
2y<3x nên 2y<9x
=>2y-9x<0
=>x-2y=0
=>x=2y
\(A=\dfrac{3x^2+xy-2y^2}{3x^2+5xy+2y^2}=\dfrac{3\cdot\left(2y\right)^2+2y\cdot y-2y^2}{3\cdot\left(2y\right)^2+5\cdot2y\cdot y+2y^2}\)
\(=\dfrac{12y^2+2y^2-2y^2}{12y^2+10y^2+2y^2}=\dfrac{12}{24}=\dfrac{1}{2}\)
\(9x^2+4y^2=20xy\Rightarrow9\left(\dfrac{x}{y}\right)^2+4=\dfrac{20x}{y}\) (1)
Đặt \(\dfrac{x}{y}=t\), do \(2y< 3x\Rightarrow\dfrac{x}{y}>\dfrac{2}{3}\Rightarrow t>\dfrac{2}{3}\)
(1) trở thành \(9t^2+4=20t\Rightarrow9t^2-20t+4=0\)
\(\Rightarrow9t^2-18t-2t+4=0\)
\(\Rightarrow9t\left(t-2\right)-2\left(t-2\right)=0\)
\(\Rightarrow\left(9t-2\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{2}{9}< \dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{3\left(\dfrac{x}{y}\right)^2+\dfrac{x}{y}-2}{3\left(\dfrac{x}{y}\right)^2+\dfrac{5x}{y}+2}=\dfrac{3.2^2+2-2}{3.2^2+5.2+2}=...\)
\(P=\left(\dfrac{2x}{3x+1}-1\right):\left(1-\dfrac{8x^2}{9x^2-1}\right)\)
Rút gọn P
Tìm x nguyên để P nguyên
a: ĐKXĐ: \(x\notin\left\{1;-1;\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
\(P=\left(\dfrac{2x}{3x+1}-1\right):\left(1-\dfrac{8x^2}{9x^2-1}\right)\)
\(=\dfrac{2x-3x-1}{3x+1}:\dfrac{9x^2-8x^2-1}{9x^2-1}\)
\(=\dfrac{-x-1}{3x+1}\cdot\dfrac{9x^2-1}{x^2-1}\)
\(=\dfrac{-\left(x+1\right)}{3x+1}\cdot\dfrac{\left(3x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{-3x+1}{x-1}\)
b: Để P là số nguyên thì \(-3x+1⋮x-1\)
=>\(-3x+3-2⋮x-1\)
=>\(-2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)
x + 2/5 = 2 – 3/7 mọi người giúp em với ạ
\(x+\dfrac{2}{5}=2-\dfrac{3}{7}\)
=>\(x+\dfrac{2}{5}=\dfrac{11}{7}\)
=>\(x=\dfrac{11}{7}-\dfrac{2}{5}=\dfrac{55-14}{35}=\dfrac{41}{35}\)