26.
februārī
Diagnosticējošais darbs MATEMĀTIKĀ 6. KLASEI
Trenējies ŠEIT!

### Teorija

Svarīgi!
${\left(a-b\right)\left(a-b\right)=\left(a-b\right)}^{2}={a}^{2}-2\mathit{ab}+{b}^{2}$

"Divu skaitļu starpības kvadrāts vienāds ar pirmā skaitļa kvadrātu mīnus divkāršots abu skaitļu reizinājums, plus
Formulas izvedums:
$\begin{array}{l}{\left(a-b\right)}^{2}=\left(a-b\right)\cdot \left(a-b\right)=\\ =a\cdot a+a\cdot \left(-b\right)-b\cdot a-b\cdot \left(-b\right)=\\ ={a}^{2}-\underset{¯}{\mathit{ab}}-\underset{¯}{\mathit{ba}}+{b}^{2}=\\ ={a}^{2}-2\mathit{ab}+{b}^{2}\end{array}$

Piemērs:
Formulas pielietojums:
${\left(a-b\right)}^{2}={a}^{2}-2\mathit{ab}+{b}^{2}$

1)
$\begin{array}{l}{\left(x-3\right)}^{2}={x}^{2}-2\cdot x\cdot 3+{3}^{2}=\\ ={x}^{2}-6x+9\end{array}$

2)
$\begin{array}{l}{\left(4x-y\right)}^{2}={\left(4x\right)}^{2}-2\cdot 4x\cdot y+{y}^{2}=\\ =16{x}^{2}-8\mathit{xy}+{y}^{2}\end{array}$

3)
$\begin{array}{l}{\left(6z-9\right)}^{2}={\left(6z\right)}^{2}-2\cdot 6z\cdot 9+{9}^{2}=\\ ={36z}^{2}-108z+81\end{array}$

Iekavu atvēršana bez formulām (reizinot kā polinomus - pirmo iekavu ar otro iekavu).

1)
$\begin{array}{l}{\left(x-3\right)}^{2}=\left(x-3\right)\cdot \left(x-3\right)=\\ =x\cdot x+x\cdot \left(-3\right)-3\cdot x-3\cdot \left(-3\right)=\\ ={x}^{2}-3x-3x+9={x}^{2}-6x+9\end{array}$

2)
$\begin{array}{l}{\left(4x-y\right)}^{2}=\left(4x-y\right)\cdot \left(4x-y\right)=\\ =4x\cdot 4x+4x\cdot \left(-y\right)-y\cdot 4x-y\cdot \left(-y\right)=\\ =16{x}^{2}-4\mathit{xy}-4\mathit{xy}+{y}^{2}=\\ =16{x}^{2}-8\mathit{xy}+{y}^{2}\end{array}$

3)
$\begin{array}{l}{\left(6z-9\right)}^{2}=\left(6z-9\right)\cdot \left(6z-9\right)=\\ =6z\cdot 6z+6z\cdot \left(-9\right)-9\cdot 6z-9\cdot \left(-9\right)=\\ =36{z}^{2}-54z-54z+81=\\ =36{z}^{2}-108z+81\end{array}$
Atsauce:
Matemātika 7. klasei /Ilze France, Gunta Lāce, Ligita Pickaine, Anita Miķelsone - Lielvārde: Lielvārds, 2007. -144.lpp.