Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH - Hỏi đáp

a)\(pt\Leftrightarrow\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)=0\)

b)\(pt\Leftrightarrow\left(x^2-x-3\right)\left(x^2+x-1\right)=0\)

ĐKXĐ: \(-1\le x\le8\) Đặt \(t=\sqrt{x+1}+\sqrt{8-x}\) ( Với \(t\ge0\))

\(\Rightarrow t^2=9+2\sqrt{\left(x+1\right)\left(8-x\right)}\)\(\Rightarrow\sqrt{\left(x+1\right)\left(8-x\right)}=\dfrac{t^2-9}{2}\)

\(\Rightarrow t+\dfrac{t^2-9}{2}=3\Rightarrow t^2+2t-15=0\)\(\Rightarrow\left(t+5\right)\left(t-3\right)=0\)

\(\left[{}\begin{matrix}t=-5\left(Loai\right)\\t=3\end{matrix}\right.\Rightarrow t=3\)

\(\Rightarrow3+\sqrt{\left(x+1\right)\left(8-x\right)}=3\) \(\Rightarrow\sqrt{\left(x+1\right)\left(8-x\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\) Thỏa mãn điều kiện .

\(C=x^2+2y^2-2xy-4y+5=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)

Đẳng thức xảy ra khi x = y = 2

Vậy min C = 1 khi x = y = 2

Ta có : C = (x2 - 2xy + y2) + ( y2 – 4y+4)+1 = (x –y)2 + (y -2)2 + 1 Vì (x – y)2 ≥ 0 ; (y-2)2 ≥ 0 Do vậy: C ≥ 1 với mọi x;y Dấu “ = ” Xảy ra khi x-y = 0 và y-2 =0 ⇔ x=y =2Vậy: Min C = 1 khi x = y =2
 

\(\Leftrightarrow\left\{{}\begin{matrix}A=0\\B=0\end{matrix}\right.\)

a/ \(x^2-25=0\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=5^2\\x^2=\left(-5\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy ...

b/ \(\left(2x-3\right)^2-\left(2x-1\right)^2=0\)

\(\Leftrightarrow\left(2x-3\right)^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-1\\2x-3=-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-2x=3-1\\2x+2x=1+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=2\left(loại\right)\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

Vậy ...

Ta có: \(P\left(x\right)=x^5+ax^4+bx^3+cx^2+dx+e\)

Suy ra \(P\left(1\right)=1^5+a\cdot1^4+b\cdot1^3+c\cdot1^2+d\cdot1+e=1\)

\(\Rightarrow a+b+c+d+e=0\)

\(P\left(2\right)=2^5+a\cdot2^4+b\cdot2^3+c\cdot2^2+d\cdot2+e=4\)

\(\Rightarrow16a+8b+4c+2d+e+28=0\)

\(P\left(3\right)=3^5+a\cdot3^4+b\cdot3^3+c\cdot3^2+d\cdot3+e=9\)

\(\Rightarrow81a+27b+9c+3d+e+234=0\)

\(P\left(4\right)=4^5+a\cdot4^4+b\cdot4^3+c\cdot4^2+d\cdot4+e=16\)

\(\Rightarrow256a+64b+16c+4d+e+1008=0\)

\(P\left(5\right)=5^5+a\cdot5^4+b\cdot5^3+c\cdot5^2+d\cdot5+e=25\)

\(\Rightarrow625a+125b+25c+5d+e+999=0\)

Thay lẫn lộn vào nhau đi nhé

Cho phép lm tiếp....

\(\Rightarrow\left\{{}\begin{matrix}15a+7b+3c+d=-28\\80a+26b+8c+2d=-234\\255a+63b+15c+3d=-1008\\624a+124b+24c+4d=-3100\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}50a-12b+2c=-178\\210a+42b+6c=-924\\564a+96b+12c=-2988\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=85\\c=-224\end{matrix}\right.\)

Thay bào pt \(15a+7b+3c+d=-28\) ta có: \(-225+595-672+d=-28\Rightarrow d=274\)

Thay vào pt \(a+b+c+d+e=0\) ta có:

\(-15+85-224+274+e=0\Rightarrow e=-120\)

Thay a,b,c,d,e vào r` tính là ra!

p/s: cho a,b,c bấm casio nhé!

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