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Tìm x:
b.2x3+4x=0 d.(x+1)=(x+1)3
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\(x\left(x-\frac{1}{3}\right)< 0\)
Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau
Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)
(4x-12)(x3+64)=0
=> [x3+64=0=>x=4x-12=0=>4x=12=>x=3 olm bị lỗi nên em đừng có viết cách ra 1 quãng như kia nhé !
vậy x thuộc {3;4}
(3x-12)(x2-4)=0
=>[x2-4=0=>x2=4=>x=2 hoặc x=-23x-12=0=>3x=12=>x=4
vậy x thuộc {4;2;-2}
(x+3)3:3-1=-10
(x+3)3:3=-9
(x+3)3=-9.3
=>(x+3)3=-27
=>x+3=-3
=>x=-6
(3x-1)3-2=-66
(3x-1)3=-64
(3x-1)3=-43
=>3x-1=-4
=>3x=-3
=>x=-1
\(\left(4x-12\right)\left(x^3+64\right)=0\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=0+12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12\div4\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow x^3+64=0\)
\(\Leftrightarrow x^3=0=64\)
\(\Leftrightarrow x^3=\left(-64\right)\)
\(\Leftrightarrow x^3=\left(-4\right)^3\)
\(\Leftrightarrow x=\left(-4\right)\)
\(\Rightarrow x\in\left\{-4;3\right\}\)
\(\Leftrightarrow\left(3x-12\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=0+12\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=12\div3\)
\(x=4\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=0+4\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x\in\left\{2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;2;4\right\}\)
Các câu khác tương tự nhé !
\(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
Ta có: \(\hept{\begin{cases}\left|x-1\right|\ge0\forall x\\\left|y+2\right|\ge0\forall x\\\left|z-3\right|\ge0\forall x\end{cases}\Rightarrow\left|x-1\right|+\left|y+2\right|+\left|z-3\right|\ge0\forall x;y;z}\)
Mà \(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
\(\hept{\begin{cases}\left|x-1\right|=0\\\left|y+2\right|=0\\\left|z-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=3\end{cases}}\)
Vậy \(x=1;y=-2;z=3\)
Lời giải:
a.
\(C(x)=A(x)+B(x)=(2x^3-3x^2-x+1)+(-2x^3+3x^2+5x-2)\)
\(=(2x^3-2x^3)+(-3x^2+3x^2)+(-x+5x)+(1-2)=4x-1\)
b.
$C(x)=4x-1=0$
$\Rightarrow x=\frac{1}{4}$
Vậy $x=\frac{1}{4}$ là nghiệm của $C(x)$
c.
\(D(x)=A(x)-B(x)=(2x^3-3x^2-x+1)-(-2x^3+3x^2+5x-2)\)
\(=2x^3-3x^2-x+1+2x^3-3x^2-5x+2\)
\(=4x^3-6x^2-6x+3\)
Bài 4.
\(A=2x^3+(x+1)^3-3x(x-2)(x+2)-3(x^2+5x+9)\\=2x^3+(x^3+3x^2+3x+1)-3x(x^2-4)-3x^2-15x-27\\=2x^3+x^3+3x^2+3x+1-3x^3+12x-3x^2-15x-27\\=(2x^3+x^3-3x^3)+(3x^2-3x^2)+(3x+12x-15x)+(1-27)\\=-26\\---\)
\(B=x(x-4x)+x(2-x)(x+2)+4(2x^2-5x+4)\\=x\cdot(-3x)+x(2-x)(2+x)+8x^2-20x+16\\=-3x^2+x(4-x^2)+8x^2-20x+16\\=-3x^2+4x-x^3+8x^2-20x+16\)
Bạn kiểm tra lại đề giúp mình!
\(C=(x-2y)(x^2+2xy+4y^2)-(x^3-8y^3+10)\) (sửa đề)
\(=x^3-(2y)^3-x^3+8y^2-10\\=x^3-8y^3-x^3+8y^3-10\\=(x^3-x^3)+(-8y^3+8y^3)-10\\=-10\)
Bài 5.
\(d)xy^2-3x^3y^2-2x(xy-3xy^2)\\=xy^2-3x^3y^2-2x^2y+6x^2y^2\\---\\f)(x-y)(2x+y)-2x^2+y^2+3xy\\=x(2x+y)-y(2x+y)-2x^2+y^2+3xy\\=2x^2+xy-2xy-y^2-2x^2+y^2+3xy\\=(2x^2-2x^2)+(xy-2xy+3xy)+(-y^2+y^2)\\=2xy\)
\(Toru\)
\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
\(=\left(x^3-2x^2+x+2x^2-4x+2-2x+7\right):\left(x^2-2x+1\right)\\ =\left[\left(x^2-2x+1\right)\left(x+2\right)-2x+7\right]:\left(x^2-2x+1\right)\\ =x+2\left(dư:-2x+7\right)\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{9}{14}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{4}{7}\)
\(x=\dfrac{19}{21}\)
a) 2x3 + 4x = 0
=> 2x(x2 + 4) = 0
=> \(\orbr{\begin{cases}2x=0\\x^2+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^2=-4\left(vl\right)\end{cases}}\)
d) \(\left(x+1\right)=\left(x+1\right)^3\)
=> \(\left(x+1\right)\left[1-\left(x+1\right)^2\right]=0\)
=> \(\orbr{\begin{cases}x+1=0\\1-\left(x+1\right)^2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\\left(x+1\right)^2=1\end{cases}}\)
=> x = -1
hoặc x + 1 = 1 hoặc x + 1 = -1
=> x = -1
hoặc x = 0 hoặc x = -2
\(b,2x^3+4x=0\)
\(\Rightarrow2x\left(x^2+2\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x=0\\x^2+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-2\left(ktm\right)\end{cases}}}\)
Vậy x=0
\(d,\left(x+1\right)=\left(x+1\right)^3\)
\(\Rightarrow\left(x+1\right)\left[1-\left(x+1\right)^2\right]=0\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)x=0\)
\(\Rightarrow x=0\) hoặc x+1=0 hoặc x+2=0
\(\Rightarrow x=0\) hoặc x=-1 hoặc x=-2
Vây \(x\in\left\{0;-1;-2\right\}\)