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1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
A . 3x + 2(x + 1) = 6x - 7
<=> 3x + 2x + 2 = 6x -7
<=> 5x - 6x = -7 - 2
<=> -x = -9
<=> x =9
B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)
=> 3(x +3) < 5(5 -x)
<=> 3x+9 < 25 - 5x
<=> 3x + 5x < 25 - 9
<=> 8x < 16
<=> x < 2
C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)
<=> 5(x - 4) + 2x = 2(x +1)
<=> 5x - 20 + 2x = 2x + 2
<=>7x - 2x = 2 + 20
<=> 5x = 22
<=> x =\(\frac{22}{5}\)
Câu 1:
Ta có phương trình: \(x^2-4x+6=\frac{21}{x^2-4x+10}\)
<=> \(\left(x^2-4x+6\right)\left(x^2-4x+10\right)=21\)
<=> \(\left(x^2-4x+8\right)^2-4=21\)
<=> \(\left(x^2-4x+8\right)^2=25\)
<=> \(x^2-4x+8=\pm5\)
<=> \(\orbr{\begin{cases}x^2-4x+3=0\\x^2-4x+13=0\end{cases}}\)
2 phương trình này bạn bấm máy tính là ra nghiệm nha :) Mình làm hơi tắt :0
Câu 3:
Ta sẽ sử dụng bất đẳng thức Bu-nhi-a-cốp-xki dạng phân thức: Với a, b, x, y thuộc R thì \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Dấu "=" xảy ra <=> \(\frac{a}{x}=\frac{b}{y}\)
Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki dạng phân thức ta có:
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
=> \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)=> đpcm
Câu 4:
Do x > 0 nên ta có: \(x+\frac{1}{x}-2=\left(\sqrt{x}\right)^2-2+\left(\frac{1}{\sqrt{x}}\right)^2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\ge0\forall x>0\)
=> \(x+\frac{1}{x}-2\ge0\Rightarrow x+\frac{1}{x}\ge2\)
=> đpcm
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{x^2-1}\)
\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)
\(\Rightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=16\)
\(\Rightarrow2\left(2x\right)=16\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
vậy \(x=4\)
\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)
\(\Rightarrow6x+1+5x-5=3x-6\)
\(\Rightarrow11x-3x=-6+4\)
\(\Rightarrow8x=-2\)
\(\Rightarrow x=\frac{-1}{4}\)
3) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\frac{x^2+x+1}{x^3-1}+\frac{\left(2x^2-5\right)}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Rightarrow3x^2-3x=-4+4\)
\(\Rightarrow3x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
ĐKXĐ: x khác 2 và -2
Ta có : \(\frac{x-2}{x+2}\)- \(\frac{x+2}{x-2}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)- \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)
<=> \(\frac{2x.\left(-4\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)
<=> -40x= -24(x^2-4)
<=> -40x= -24x^2+96
<=> 24x^2-40x-96=0
<=> 24x^2-72x+32x-96=0
<=> 24x(x-3)+32(x-3)=0
<=> (x-3)(24x+32)=0
=> \(\orbr{\begin{cases}x-3=0\\24x+32=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{-4}{3}\end{cases}}\)
Vậy S=\(\hept{\begin{cases}\\\end{cases}}3;\frac{-4}{3}\)