hình thang cân có 2 đường chéo vuông góc với nhau . Đáy nhỏ = 13,724 cm ; cạnh bên = 21,567 cm . TÍnh diện tivhs hình thang

Cho t làm bừa phát :v

\(B=\frac{15}{\sqrt{6}+1}-\frac{6}{\sqrt{6}-2}\)

\(B=\frac{15\sqrt{6}-15}{\sqrt{6^2}-1}-\frac{6\sqrt{6}+12}{\sqrt{6^2}-2^2}\)

\(B=\frac{15\left(\sqrt{6}-1\right)}{\sqrt{6^2}-1}-\frac{6\left(\sqrt{6}+2\right)}{\sqrt{6^2}-2^2}\)

\(B=\frac{15\left(\sqrt{6}-1\right)}{6-1}-\frac{6\left(\sqrt{6}+2\right)}{6-4}\)

\(B=\frac{15\left(\sqrt{6}-1\right)}{5}-\frac{6\left(\sqrt{6}+2\right)}{2}\)

\(B=3\left(\sqrt{6}-1\right)-3\left(\sqrt{6}+2\right)\)

\(B=3\sqrt{6}-3-3\sqrt{6}-6\)

\(B=\left(3\sqrt{6}-3\sqrt{6}\right)+\left(-3-6\right)\)

\(=-9\)

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

bạn làm như này nha:

Từ đpcm  \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)

             \(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

             \(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)      

             \(\Leftrightarrow0=2.\left(\frac{a+b+c}{abc}\right)\)

             \(\Leftrightarrow0=a+b+c\)luôn đúng do giả thuyết cho

                                \(\Rightarrowđpcm\)

A= 5 căn 3 - 3*4 căn 3 +2*5 căn 3 -1/3 *6 căn 3 = căn 3

\(A=5\sqrt{3}-3\sqrt{48}+2\sqrt{75}-\frac{1}{3}\sqrt{108}\)

\(A=5\sqrt{3}-3.\sqrt{4^2.3}+2\sqrt{5^2.3}-\frac{1}{3}.\sqrt{6^2.3}\)

\(A=5\sqrt{3}-12\sqrt{3}+10\sqrt{3}-2\sqrt{3}\)

\(A=\sqrt{3}\)

\(\text{ĐKXĐ: }x\ge0;x\ne\pm1\)

\(2\sqrt{144x+144}-3\sqrt{100x-100}=12\)

\(2\sqrt{144\left(x+1\right)}-3\sqrt{100\left(x-1\right)}=12\)

\(2\sqrt{144}.\sqrt{\left(x+1\right)}-3\sqrt{100}.\sqrt{x-1}=12\)

\(2.12\sqrt{x+1}-3.10\sqrt{x-1}=12\)

\(24\sqrt{x+1}-30\sqrt{x-1}=12\)

\(6.\left(4\sqrt{x+1}-5\sqrt{x-1}\right)=6.2\)

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

\(\text{Mk bí r}\)

\(22-2\sqrt{x+5}+1=-x^2-9x\)

\(23-2\sqrt{x+5}=-x^2-9x\)

\(2\sqrt{x+5}=23+x^2+9x\)

\(4\left(x+5\right)=\left(23+x^2+9x\right)^2\)

\(4x+20=529+x^4+81x^2+46x^2+18x^3+261x\)

\(x^4+18x^3+81x^2+257x+509=0\)

bấm máy thì ra nha

\(b,P=\sqrt{12-4\sqrt{3}}\)

\(P=\sqrt{4\left(3-\sqrt{3}\right)}\)

\(c.\sqrt{\left(2\sqrt{3}\right)^2+4\sqrt{3}+1}-\sqrt{\left(2\sqrt{3}\right)^2-4\sqrt{3}+1}\)

\(\left|2\sqrt{3}+1\right|-\left|2\sqrt{3}-1\right|\)

\(2\sqrt{3}+1-2\sqrt{3}+1\)

\(=2\)