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\(a,=\left(\sqrt{6}+\sqrt{5}\right)^2\\ b,=\left(\sqrt{7}-\sqrt{3}\right)^2\\ c,=\left(\sqrt{3}+\sqrt{5}\right)^2\\ d,=\left(2+\sqrt{3}\right)^2\\ e,=\left(2\sqrt{2}-1\right)^2\)
a: \(=\left(\sqrt{6}+\sqrt{5}\right)^2\)
b: \(=\left(\sqrt{7}-\sqrt{3}\right)^2\)
b: \(10-2\sqrt{21}=\left(\sqrt{7}-\sqrt{3}\right)^2\)
c: \(8+2\sqrt{15}=\left(\sqrt{5}+\sqrt{3}\right)^2\)
Lời giải:
a) $m^2-25=m^2-5^2=(m-5)(m+5)$
b) $k^2-7=k^2-(\sqrt{7})^2=(k-\sqrt{7})(k+\sqrt{7})$
c) $3-36d^2=(\sqrt{3})^2-(6d)^2=(\sqrt{3}-6d)(\sqrt{3}+6d)$
d) $x-81=(\sqrt{x})^2-9^2=(\sqrt{x}-9)(\sqrt{x}+9)$
e) $2+5a=2-(-5a)=(\sqrt{2})^2-(\sqrt{-5a})^2=(\sqrt{2}-\sqrt{-5a})(\sqrt{2}+\sqrt{-5a})$
a.
ĐKXĐ: $x\geq 0; y\geq 1$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+(y-1-6\sqrt{y-1}+9)=0$
$\Leftrightarrow (\sqrt{x}-2)^2+(\sqrt{y-1}-3)^2=0$
Vì $(\sqrt{x}-2)^2; (\sqrt{y-1}-3)^2\geq 0$ với mọi $x\geq 0; y\geq 1$ nên để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{y-1}-3=0$
$\Leftrightarrow x=4; y=10$
b.
ĐKXĐ: $x\geq -1; y\geq -2; z\geq -3$
PT $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$
$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$
$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$
b: \(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
c: \(13+\sqrt{48}=13+4\sqrt{3}=\left(2\sqrt{3}+1\right)^2\)
d: \(4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(a,8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}+1^2=\left(\sqrt{7}-1\right)^2\)
\(b,8-2\sqrt{15}=\sqrt{5}^2-2.\sqrt{3}.\sqrt{5}+\sqrt{3}^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(c,8+4\sqrt{3}=2^2+2.2.\sqrt{3}+\sqrt{3}^2=\left(2+\sqrt{3}\right)^2\)
Ta có:
\(a^2+b^2+c^2+d^2=\left(b+c+d\right)^2+b^2+c^2+d^2\)
\(=2\left(b^2+c^2+d^2\right)+2\left(bd+cd+bc\right)\)
\(=\left(b^2+2bc+c^2\right)+\left(c^2+2cd+d^2\right)+\left(d^2+2db+b^2\right)\)
\(=\left(b+c\right)^2+\left(c+d\right)^2+\left(d+b\right)^2\)
đề có a mà sao kết quả lại mất a vậy