\(\dfrac{5+4x}{7x}+\dfrac{x-3}{7x}\)
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a: \(=2x+x^3-5x^4\)
b: \(=\dfrac{8x^2+4x-7x-3}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x^2-3x-3}{\left(2x-1\right)\left(2x+1\right)}\)
\(\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{x}{2x+3}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\) ( sửa đề )
\(=\left[\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\right].\dfrac{x}{2x+3}\)
\(=\dfrac{x}{2x+3}\)
a: =>x^2+4x+4-x^2+4x-4>8x-2
=>8x>8x-2
=>0>-2(luôn đúng)
b: =>15x-10(7x+5)-24x>-240
=>15x-70x-50-24x>-240
=>-79x-50>-240
=>-79x>-190
=>x<190/79
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)
\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)
\(\Leftrightarrow9x-\left|7x+5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)
\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)
\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)
\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)
\(\Leftrightarrow-5=21-8\sqrt{x+5}\)
\(\Leftrightarrow8\sqrt{x+5}=21+5\)
\(\Leftrightarrow8\sqrt{x+5}=26\)
\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)
\(\Leftrightarrow x+5=\dfrac{169}{16}\)
\(\Leftrightarrow x=\dfrac{169}{16}-5\)
\(\Leftrightarrow x=\dfrac{89}{16}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)
Nick cũ không đi giải lấy nick mới giải làm gì vậy Tuấn Anh Phan Nguyễn ? :D
\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{x^3-x^3+3x^2-3x+1}{\left(4x+3\right)\left(x-5\right)}=\dfrac{\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)}{\left(4x+3\right)\left(x-5\right)}\)
\(\Leftrightarrow3x^2-3x+1=7x^2-35x-x+5-4x^2-3x\)
\(\Leftrightarrow3x^2-3x+1=3x^2-39x+5\)
\(\Leftrightarrow-3x+1=-39x+5\)
\(\Leftrightarrow36x=4\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy...
\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\) ( đk: x ≠ \(\dfrac{-3}{4}\) ; x ≠ 5 )
\(\Leftrightarrow\) \(x^3-\left(x-1\right)^3=\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)\)
\(\Leftrightarrow\) \(\left(x-x+1\right)\left(x^2+x^2-x+x^2-2x+1\right)=7x^2-35x-x+5-4x^2-3x\)\(\Leftrightarrow\) \(3x^2-3x+1=3x^2-39x+5\)
\(\Leftrightarrow\) \(36x=4\)
\(\Leftrightarrow\) \(x=\dfrac{1}{9}\)( TM )
\(S=\left\{\dfrac{1}{9}\right\}\)
a: \(\Leftrightarrow15\left(x-1\right)-2\left(7x+3\right)\le10\left(2x+1\right)+6\left(3-2x\right)\)
\(\Leftrightarrow15x-15-14x-6\le20x+10+18-12x\)
=>x-21<=8x+28
=>-7x<=49
hay x>=-7
b: \(\Leftrightarrow20\left(2x+1\right)-15\left(2x^2+3\right)< 10x\left(5-3x\right)-12\left(4x+1\right)\)
\(\Leftrightarrow40x+20-30x^2-45< 50x-30x^2-48x-12\)
=>40x-25<2x-12
=>38x<13
hay x<13/38
\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)
\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
\(=\dfrac{5+4x+x-3}{7x}\left(x\ne0\right)=\dfrac{5x+2}{7x}\)
Cảm ơn nhoa