a/

\(Q\left(2\right).Q\left(-1\right)=\left(4a+2b+c\right)\left(a-b+c\right)=\left(5a+b+2c-a+b-c\right)\left(a-b+c\right)\)

\(=\left(-a+b-c\right)\left(a-b+c\right)=-\left(a-b+c\right)^2\le0\)

b/

Q(x) = 0 với mọi x, suy ra các điều sau:

\(\Rightarrow Q\left(0\right)=c=0\); \(Q\left(1\right)=a+b+c=a+b=0\); \(Q\left(-1\right)=a-b+c=a-b=0\)

\(\Rightarrow\left(a+b\right)+\left(a-b\right)=0\text{ và }\left(a+b\right)-\left(a-b\right)=0\)\(\Leftrightarrow2a=0\text{ và }2b=0\Leftrightarrow a=b=0\)

Vậy \(a=b=c=0\)

\(f\left(x\right)=ax^2+bx+c\Rightarrow\hept{\begin{cases}f\left(0\right)=c\\f\left(1\right)=a+b+c\\f\left(2\right)=4a+2b+c\end{cases}}\)

\(f\left(0\right)\) nguyên \(\Rightarrow c\) nguyên \(\Rightarrow\hept{\begin{cases}2a+2b\\4a+2b\end{cases}}\) nguyên

\(\Rightarrow\left(4a+2b\right)-\left(2a+2b\right)=2a\)(nguyên)

\(\Rightarrow2b\) nguyên

\(\Rightarrowđpcm\)

\(36-y^2\le36\)

\(8\left(x-2010\right)^2\ge0;8\left(x-2010\right)^2⋮8\)

\(\Rightarrow\hept{\begin{cases}0\le8\left(x-2010\right)^2\le36\\8\left(x-2010\right)^2⋮8\\8\left(x-2010\right)^2\in N\end{cases}}\)

Giai tiep nhe

a, 11/12 - ( 2/5 + x ) = 2/3

<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)

b, 2x . ( x - 1/7 ) = 0

<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

vậy x={\(0;\frac{1}{7}\)}

c, 3/4 + 1/4 : x = 2/5

<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)

vậy x=-5/7

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)

\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)

\(\Leftrightarrow x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow4x=\frac{-20}{7}\)

\(\Leftrightarrow x=-\frac{5}{7}\)

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{2}{5}\)

\(x=-\frac{3}{20}\)

Vậy \(x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Rightarrow\begin{cases}x=0\\x=\frac{1}{7}\end{cases}\)

Vậy x = 0 hoặc \(x=\frac{1}{7}\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)

\(\frac{1}{4}:x=-\frac{7}{20}\)

\(x=\frac{1}{4}:\left(-\frac{7}{20}\right)\)

\(x=-\frac{5}{7}\)

Vậy \(x=-\frac{5}{7}\)

Bài 1:

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=1\\2a+b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\)

Ta có: \(P\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\hept{\begin{cases}P\left(-1\right)=a\left(-1\right)^2-b+c=a-b+c\\P\left(-2\right)=a.\left(-2\right)^2-2b+c=4a-2b+c\end{cases}}\)

\(\Rightarrow P\left(-1\right).P\left(-2\right)\)

\(=\left(a-b+c\right)\left(4a-2b+c\right)\)

\(=[5a-3b+c-4a+2b-c]\left(4a-2b+c\right)\)

\(=[0-\left(4a-2b+c\right)]\left(4a-2b+c\right)\)

\(=-\left(4a-2b+c\right)\left(4a-2b+c\right)\)

\(=-\left(4a-2b+c\right)^2\)

Mặt khác \(\left(4a-2b+c\right)^2\ge0\)

\(\Rightarrow-\left(4a-2b+c\right)^2\le0\)

\(\Rightarrow P\left(-1\right).P\left(-2\right)\le0\left(đpcm\right)\)

Chắc là D:V

D. H₂SO₄

D

a) Ta có:

\(x^2+2xy+y^2+1\)

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

Vì \(\left(x+y\right)^2\ge0\) với mọi x và y

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

\(\Rightarrow\left(x+y\right)^2+1>0\) với mọi x

b) Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) với mọi x

\(\frac{1}{y}=\frac{x}{4}-\frac{1}{2}=\frac{x-2}{4}\)

Suy ra y.(x - 2) = 4. Vì x, y ∈ Z nên x - 2 ∈ Z, ta có bảng sau:
 

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