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\(a,\left|\frac{4x}{5}-\frac{2}{7}\right|-\frac{3}{2}=-\frac{2}{5}\)
\(\Leftrightarrow\left|\frac{4x}{5}-\frac{2}{7}\right|=\frac{11}{10}\)
Xét cả hai trường hợp :
Trường hợp 1 : \(\frac{4x}{5}-\frac{2}{7}=\frac{11}{10}\)
\(\Leftrightarrow\frac{4x}{5}=\frac{97}{70}\)
\(\Leftrightarrow4x=\frac{97}{14}\)
\(\Leftrightarrow x=\frac{97}{56}\)
Trường hợp 2 : \(\frac{4x}{5}-\frac{2}{7}=-\frac{11}{10}\)
\(\Leftrightarrow\frac{4x}{5}=-\frac{57}{50}\)
\(\Leftrightarrow4x=-\frac{57}{14}\)
\(\Leftrightarrow x=-\frac{57}{56}\)
\(b,\left|4x-\frac{1}{5}\right|=\left|2x+\frac{1}{2}\right|\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{5}=2x+\frac{1}{2}\\4x-\frac{1}{5}=-2x+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{5}-2x=\frac{1}{2}\\4x-\frac{1}{5}-(-2x)=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-2x-\frac{1}{5}=\frac{1}{2}\\4x-(-2x)-\frac{1}{5}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{10}\\6x=-\frac{3}{10}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=-\frac{1}{20}\end{cases}}\)
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\(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{5}=0\\x^2+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{1}{5}\\x^2=-\dfrac{1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{2}{15}\)
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ta có: f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)
\(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)
\(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)
\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)
Chúc bn học tốt !!!!!!
Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............
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a, 2.(4x-3)-3(x+5)+4(x-10)=5(x+2)
2.4x-2.3-3.x+3.5+4x-4.10=5x+5.2
8x-6-3x+15+4x-40=5x-10
8x-3x+4x-5x-6-15-40-10=0
4x-71=0
4x=71
x=71:4
x=71/4
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a)\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\)
\(4x^2-11x-3-\left(4x^2-29x+7\right)=15\)
\(4x^2-11x-3-4x^2+29x-7=15\)
\(18x-10=15\)
\(x=\frac{25}{18}\)
b)\(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\left(x+1\right)\left(3x-5-3x+1\right)=x-4\)
\(\left(x+1\right).\left(-4\right)-x+4=0\)
\(-4x-4-x+4=0\)
\(x=0\)
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a,2x-3=x+1/2 b,4x-(x+1/2)=2x+(1/2-5) c,2/3-1/3(x-2/3)-1/2(2x+1)=5
2x-x =1/2+3 4x-x-1/2=2x+1/2-5 d,(x+1/2).(x-3/4)=0
x=7/2 4x-x-2x =1/2-5+1/2 \(\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)
x=-4
e,(2x-1)(3x+1/5)=0
\(\orbr{\begin{cases}2x-1=0\\3x+\frac{1}{5}=0\end{cases}}\orbr{\begin{cases}2x=1\\3x=\frac{1}{5}\end{cases}}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{15}\end{cases}}\)
f, 4x2-2x=0
Các câu mk chưa làm thì bạn cứ chờ để mk suy nghĩ.
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`a)`
`A(x) + B(x) = 2x - 4x^2 + 1 + x^3 - 4x^2 + 5 - 2x`
`= x^3 - ( 4x^2 + 4x^2 ) + ( 2x - 2x ) + ( 1+ 5 )`
`= x^3 - 8x^2 + 6`
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`b)`
`P(x) + B(x) = A(x)`
`=>P(x) = A(x) - B(x)`
`=>P(x) = 2x - 4x^2 + 1 + x^3 + 4x^2 - 5 + 2x`
`=>P(x) = x^3 + ( -4x^2 + 4x^2 ) + ( 2x + 2x ) + ( 1 - 5 )`
`=>P(x) = x^3 + 4x - 4`
\(\left|x^2-x\right|-\left|4x+5\right|=0,TXĐ:D=R\)
\(\Leftrightarrow\left|x^2-x\right|=\left|4x+5\right|\Leftrightarrow\orbr{\begin{cases}x^2-x=4x+5\\x^2-x=-4x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-5x-5=0\\x^2+3x+5=0\left(VN\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+3\sqrt{5}}{2}\\x=\frac{5-3\sqrt{5}}{2}\end{cases}\left(TMĐK\right)}\)
thank bạn nhưng bạn ghi rõ hộ mk đc ko mk chx học căn bậc nên ko ghi kết quả như của bạn đc