TÌm x biết
3/4 x + 1/3 = -\(\frac{1}{2}\)
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ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???
<=> \(\frac{-2x-1}{12}\)-\(\frac{2x+2}{3}=\frac{1-2x}{4}-\frac{3x-1}{12}\)
<=>\(\frac{-2x-1-8x-8-3+6x+3x-1}{12}=0\)
<=> -x-13=0=> x=-13
\(\frac{\frac{1}{2}-\frac{x+2}{3}}{2}-\frac{2}{3}\left(x+1\right)=\frac{1}{4}\left(1-2x\right)-\frac{\frac{1}{3}-\frac{1-x}{2}}{2}\)
<=>\(6.\left(\frac{1}{2}-\frac{x+2}{3}\right)-8.\left(x+1\right)=3\left(1-2x\right)-6.\left(\frac{1}{3}-\frac{1-x}{2}\right)\)
<=>3-2.(x+2)-8x-8=3-6x-2+3.(1-x)
<=>3-2x-4-8x-8=3-6x-2+3-3x
<=>-10x-9=-9x+4
<=>x=-13
\(\Leftrightarrow x+1+5x\left(x+2\right)=4\left(x-2\right)+3x^2-12\)
\(\Leftrightarrow x+1+5x^2+10x-4x+8-3x^2+12=0\)
\(\Leftrightarrow2x^2+7x+21=0\)
\(\text{Δ}=7^2-4\cdot2\cdot21=49-168< 0\)
Vì Δ<0 nên phương trình vô nghiệm
\(\Leftrightarrow3\left(x^2-4x+4\right)-\dfrac{5}{4}\left(9x^2+6x+1\right)=\dfrac{4}{3}\left(-x^2+4x-3\right)-\dfrac{7}{6}x\left(x-3\right)\)
\(\Leftrightarrow3x^2-12x+12-\dfrac{45}{4}x^2-\dfrac{15}{2}x-\dfrac{5}{4}=-\dfrac{4}{3}x^2+\dfrac{16}{3}x-4-\dfrac{7}{6}x^2+\dfrac{7}{2}x\)
\(\Leftrightarrow x^2\cdot\dfrac{-33}{4}-\dfrac{39}{2}x+\dfrac{43}{4}+\dfrac{5}{2}x^2-\dfrac{53}{6}x+4=0\)
\(\Leftrightarrow x^2\cdot\dfrac{-23}{4}-\dfrac{85}{3}x+\dfrac{59}{4}=0\)
\(\Leftrightarrow12\left(\dfrac{-23}{4}x^2-\dfrac{85}{3}x+\dfrac{59}{4}\right)=0\)
\(\Leftrightarrow-69x^2-340x+177=0\)
\(\Leftrightarrow69x^2+340x-177=0\)
\(\text{Δ}=340^2-4\cdot69\cdot\left(-177\right)=164452\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-170-\sqrt{41113}}{69}\\x_2=\dfrac{-170+\sqrt{41113}}{69}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)
\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)
\(\Leftrightarrow167x^2-255x-498=0\)
\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)
Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=\frac{2x-\frac{10-7x}{3}}{2}-\left(x+1\right)\)
<=>\(2x-\frac{x}{2}+\frac{3+x}{4}=2x-\frac{10-7x}{3}-2\left(x+1\right)\)
<=>\(24x-6x+9+3x=24x-40+28x-24x-24\)
<=>\(21x+9=28x-64\)
<=>\(-7x=-73\)
<=>x=73/7
3-(3/4+x-1/3):2/3=1/2
3-(3/4+x-1/3)=1/2*2/3
3-(3/4+x-1/3)=2/6
3/4+x-1/3=3-2/6
3/4+x-1/3=8/3
3/4+x=8/3+1/3
3/4+x=3
x=3-3/4
x=9/4