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5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
Chúc bn học tốt❤
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\left(đkxđ:x\ne-4;-5;-6;-7\right)\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-13\left(tm\right)\end{matrix}\right.\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\\ ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\\ \Rightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{x+5}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{18\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}-\dfrac{18\left(x+5\right)}{18\left(x+5\right)\left(x+7\right)}=\dfrac{\left(x+5\right)\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}\\ \Rightarrow18x+126-18x-90=x^2+5x+7x+35\\ \Leftrightarrow x^2+12x+35=36\\ \Leftrightarrow x^2+12x-1=0\\ \Leftrightarrow x^2+12x+36-37=0\\ \Leftrightarrow\left(x^2+12x+36\right)-37=0\\ \Leftrightarrow\left(x+6\right)^2-37=0\\ \Leftrightarrow\left(x+6+\sqrt{37}\right)\left(x+6-\sqrt{37}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+6+\sqrt{37}=0\\x+6-\sqrt{37}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6-\sqrt{37}\\x=\sqrt{37}-6\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\sqrt{37}-6;-\sqrt{37}-6\right\}\)
Câu 1:
\(\left(x+\dfrac{2}{3}\right)\cdot\left(x-\dfrac{1}{2}\right)=0\)
=>\(\left[{}\begin{matrix}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Câu 2:
x+1=2x+3
=>x-2x=3-1
=>-x=2
=>x=-2
=>-2 là nghiệm
Câu 3:
ĐKXĐ: x<>-5
\(\dfrac{\left(-x+2\right)\left(2x+10\right)}{x^2+10x+25}=0\)
=>\(\dfrac{\left(-x+2\right)\cdot2\cdot\left(x+5\right)}{\left(x+5\right)^2}=0\)
=>\(\dfrac{2\left(-x+2\right)}{\left(x+5\right)}=0\)
=>-x+2=0
=>x=2(nhận)
Câu 4:
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)
Câu 10: ĐKXĐ: x<>1
\(x^2+\dfrac{1}{x-1}=1+\dfrac{1}{1-x}\)
=>\(x^2-1+\dfrac{1}{x-1}+\dfrac{1}{x-1}=0\)
=>\(\left(x-1\right)\left(x+1\right)+\dfrac{2}{x-1}=0\)
=>\(\dfrac{\left(x^2-1\right)\cdot\left(x-1\right)+2}{x-1}=0\)
=>\(x^3-x^2-x+1+2=0\)
=>\(x^3-x^2-x+3=0\)
=>\(x\simeq-1,36\)
\(a.\dfrac{2x-1}{x-1}+\dfrac{x}{x^2-3x+2}=\dfrac{6x-2}{x-2}\left(x\ne2;x\ne1\right)\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-2\right)+x}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(6x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2-4x-x+2+x=6x^2-6x-2x+2\)
\(\Leftrightarrow2x^2-5x+2=6x^2-8x+2\)
\(\Leftrightarrow4x^2-3x=0\)
\(\Leftrightarrow x\left(4x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{3}{4}\left(TM\right)\end{matrix}\right.\)
KL........
\(b.A=\sqrt{x^2-x+1\dfrac{1}{4}}-2016=\sqrt{x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+1}-2016=\sqrt{\left(x-\dfrac{1}{2}\right)^2+1}-2016\ge1-2016=-2015\)
\(\Rightarrow A_{Min}=-2015."="\Leftrightarrow x=\dfrac{1}{2}\)
Đề sai thì phải, bạn thêm dấu ngoặc vào đi. Như vậy dễ làm hơn.
bạn ấn vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm
ĐKXĐ: \(x\ne-1;x\ne2\)
\(\dfrac{x^2-x}{x^2-x+1}-\dfrac{x^2-x+2}{x^2-x-2}=1\) (1)
\(\Leftrightarrow\dfrac{x^2-x}{x^2-x+1}-\dfrac{x^2-x+2}{x^2-x-2}-1=0\)
\(\Leftrightarrow\dfrac{\left(x^2-x-2\right)\left(x^2-x\right)-\left(x^2-x+1\right)\left(x^2-x+2\right)-\left(x^2-x+1\right)\left(x^2-x-2\right)}{\left(x^2-x+1\right)\left(x^2-x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^3-5x^2+4x-x^4}{\left(x^2-x+1\right)\left(x^2-x-2\right)}=0\)
\(\Leftrightarrow2x^3-5x^2+4x-x^4=0\)
\(\Leftrightarrow x\left(2x^2-5x+4-x^3\right)=0\)
\(\Leftrightarrow x\left(-x^3+2x^2-5x+4\right)=0\)
\(\Leftrightarrow x\left(-x^3+x^2+x^2-x-4x+4\right)=0\)
\(\Leftrightarrow x\left[-\left(x-1\right)\right]\left(x^2-x+4\right)=0\)
\(\Leftrightarrow-x\left(x-1\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x-1=0\\x^2-x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(đk:x\ne-1;x\ne2\right)\\x\notin R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{0;1\right\}\)