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`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)
`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`
`C(x)= -3x^4-2x^3+x^2+x+5`
`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)
`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`
`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`
`D(x)=-x^4-4x^3-4x+7`
`P(x)=C(x)+D(x)`
`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`
`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`
`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`
`P(x)=-4x^4-6x^3+x^2-3x+12`
`Q(x)=C(x)-D(x)`
`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`
`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`
`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`
`Q(x)=-2x^4+2x^3+x^2+5x-2`
`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`
`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`
`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`
`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`
`F(x)=5x+10`
Đặt `5x+10=0`
`\Leftrightarrow 5x=0-10`
`\Leftrightarrow 5x=-10`
`\Leftrightarrow x=-10 \div 5`
`\Leftrightarrow x=-2`
Vậy, nghiệm của đa thức là `x=-2.`
a: Ta có: \(3\left|2x+5\right|\ge0\forall x\)
\(\Leftrightarrow3\left|2x+5\right|-7\ge-7\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)
c: ta có: \(\left(2x-3\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(2x-3\right)^2-14\ge-14\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp
c) l x - 5 l = 2x
\(\Leftrightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2x=5\\x+2x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)
Hok tốt!!!!!!!
Tìm x, biết:
a) |2x + 1| = 17
<=>\(\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\)
<=> \(\hept{\begin{cases}x=8\\x=-9\end{cases}}\)
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
a: \(\left|7-2x\right|+7=2x\)
=>\(\left|2x-7\right|+7=2x\)
=>\(\left|2x-7\right|=2x-7\)
=>2x-7>=0
=>\(x>=\dfrac{7}{2}\)
b: \(\left|1-x\right|=4x+1\)
=>\(\left|x-1\right|=4x+1\)
=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)
d: \(\left|x-7\right|+2x+5=6\)
=>\(\left|x-7\right|=6-2x-5=-2x+1\)
=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
e: 3x-|2x-1|=2
=>|2x-1|=3x-2
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
2x + 3 = 5
=> 2x = 2
=> x = 1
giải
th1
2x+3=5
2x =2
x = 1
th2
2x+3=-5
2x =-5 -3
2x = -8
x = -4