a, \(AC=\dfrac{AB}{sin45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)

\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}=a.a\sqrt{2}.cos45^o=a^2\)

b, \(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)

\(=\overrightarrow{AC}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BC}\)

\(=AC.BD.cos90^o+AC.AD.cos45^o\)

\(=a\sqrt{2}.a\sqrt{2}.0+a\sqrt{2}.a.\dfrac{\sqrt{2}}{2}=a^2\)

c, \(\overrightarrow{AB}.\overrightarrow{BD}=AB.BD.cos135^o=-a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2\)

d, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\left(2\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{BC}.\left(\overrightarrow{AD}+\overrightarrow{BD}\right)\)

\(=\overrightarrow{BC}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BD}\)

\(=AD^2+BC.BD.cos45^o\)

\(=a^2+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=2a^2\)

e, \(\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right)\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\right)\)

\(=\left(\overrightarrow{AC}+\overrightarrow{AC}\right)\left(\overrightarrow{DB}+\overrightarrow{DB}\right)\)

\(=4.\overrightarrow{AC}.\overrightarrow{DB}=4.AC.DB.cos90^o=0\)

1.

\(\overrightarrow{AB}.\overrightarrow{BC}=\overrightarrow{AB}.\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}.\left(-\overrightarrow{AB}\right)+\overrightarrow{AB}.\overrightarrow{AC}=-AB^2=-25\)

2.

\(\overrightarrow{AB}.\overrightarrow{BD}=\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AD}\right)=-\overrightarrow{AB}.\overrightarrow{AB}+\overrightarrow{AB}.\overrightarrow{AD}=-AB^2+0=-64\)

cứu

Câu 4:

a: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\forall a,b\) thỏa mãn ĐKXĐ

=>\(a-2\sqrt{ab}+b\ge0\forall a,b\) thỏa mãn ĐKXĐ

=>\(a+b\ge2\sqrt{ab}\forall a,b\) thỏa mãn ĐKXĐ

=>\(\frac{a+b}{2}\ge\sqrt{ab}\forall a,b\) thỏa mãn ĐKXĐ

Câu 2:

a: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+b^2d^2+2\cdot acbd+a^2d^2+b^2c^2-2\cdot ad\cdot bc\)

\(=a^2c^2+b^2c^2+b^2d^2+a^2d^2\)

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

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

b: \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

=>\(a^2c^2+b^2d^2+2\cdot acbd\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

=>\(a^2d^2+b^2c^2\ge2abcd\)

=>\(a^2d^2-2\cdot ad\cdot bc+b^2c^2\ge0\)

=>\(\left(ad-bc\right)^2\ge0\forall a,b,c,d\) (luôn đúng)

Câu 1: Giả sử \(\sqrt7\) là số hữu tỉ

=>\(\sqrt7=\frac{a}{b}\) , với ƯCLN(a;b)=1

=>\(\left(\frac{a}{b}\right)^2=7\)

=>\(a^2=7b^2\)

=>\(a^2\) ⋮7

=>a⋮7

=>a=7k

\(a^2=7b^2\)

=>\(7b^2=\left(7k\right)^2=49k^2\)

=>\(b^2=7k^2\) ⋮7

=>b⋮7

=>ƯCLN(a;b)=7, khác với giả sử

=>\(\sqrt7\) là số vô tỉ