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a. \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)
\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)
\(\Rightarrow x=1;y=-2;z=-1\)
<=>(x2-2x+1)+(3y2+12y+12)+(2z2+4z+2)=0
<=>(x-1)2+3(y+2)2+2(z+1)2=0
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\3\left(y+2\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)
1)
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-19\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2+3x^2-6x^2\right)+\left(3x-3x+12x\right)+\left(1+1-6+19\right)=0\)
\(\Leftrightarrow12x+15=0\)
\(\Leftrightarrow x=-\frac{5}{4}\)
Bài 1:
Ta có:
\(2x^2+4x^3-7=4x^2(x-3)+14x(x-3)+42(x-3)+119\)
\(=(x-3)(4x^2+14x+42)+119\)
Do đó phép chia $2x^2+4x^3-7$ cho $x-3$ có thương là $4x^2+14x+42$ và dư là $119$
Bài 2:
Theo định lý Bê-du về phép chia đa thức thì phép chia đa thức $f(x)$ cho $x-a$ có dư là $f(a)$
Áp dụng vào bài toán:
\(f(2)=-23\)
\(\Leftrightarrow 2^3-4.2^2+5.2+a=-23\)
\(\Leftrightarrow 2+a=-23\Rightarrow a=-25\)
Bài 3:
Ta có:
\(x^3+ax+b=x(x^2+2x+1)-2x^2-x+ax+b\)
\(=x(x^2+2x+1)-2(x^2+2x+1)+3x+2+ax+b\)
\(=(x-2)(x+1)^2+x(a+3)+(b+2)\)
Vậy $x^3+ax+b$ khi chia $(x+1)^2$ có dư là $x(a+3)+(b+2)$
\(\Rightarrow \left\{\begin{matrix} a+3=2\\ b+2=1\end{matrix}\right.\Rightarrow a=-1; b=-1\)
Bài 4:
\(x^2+y^2-4y+5=0\)
\(\Leftrightarrow x^2+(y^2-4y+4)+1=0\)
\(\Leftrightarrow x^2+(y-2)^2+1=0\)
\(\Rightarrow x^2+(y-2)^2=-1\)
Rõ ràng vế trái luôn không âm, mà vế phải âm nên vô lý
Vậy pt vô nghiệm, không tồn tại $x,y$ thỏa mãn.
a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
Bài 1:
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow x^2-2x+1+3y^2+12y+12+2z^2+4z+2=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
Dễ thấy: \(\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\3\left(y+2\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
Bài 2:
a)\(A=x^2-4xy+5y^2+10x-22y+28\)
\(=x^2-4xy+10x+4y^2-20y+25+y^2-2y+1+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
b)\(B=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+15\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+15\)
Đặt \(t=x^2-5x+4\) thì ta có:
\(t\left(t+2\right)+15=t^2+2t+1+14\)
\(=\left(t+1\right)^2+14\ge14\)
Xảy ra khi \(t=-1 \)\(\Rightarrow x^2-5x+4=-1\Rightarrow x=\dfrac{5\pm\sqrt{5}}{2}\)
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
Chia nhỏ ra bạn ơi!
\(a) x² +3y²+2z²-2x+12y+4z+15=0 \)
\(⇔x²-2x+1+3y²+12y+12+2z²+4z+2=0 \)
\(⇔(x²-2x+1) + 3(y²+4y+4) +2(z²+2z+1)=0 \)
\(⇔(x-1)² +3(y+2)²+2(z+1)²=0 \)
\(⇔ x-1=0 \) và \(y+2=0\) và \(z+1=0\)
Vậy: \(x=1;y=-2;z=-1\)