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\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
\(=-5\)
\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=\sqrt{5}.\left(5-12+6-4\right)\)
\(=-5\sqrt{5}\)
\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)
\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)
\(=2.3+5-5.2\)
\(=1\)
\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)
\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)
\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)
\(=0\)
\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)
\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
L=2*3+5-5*2=5-4=1
N=8căn 2-8căn2-15căn3+15căn 3=0
O=10căn 3-6căn3+3căn3=7căn 3
Bài 4:
Để hai đường song song thì 2m-1=5
=>2m=6
=>m=3
Bài 3:
a: 4x-3y=2 và 4x+3y=-18
=>8x=-16 và 4x-3y=2
=>x=-2 và 3y=4x-2=4*(-2)-2=-10
=>x=-2; y=-10/3
b:\(A=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\left(x+16\right)^2}{\left(x-16\right)\left(\sqrt{x}+2\right)}\)
\(2x^2+3x-5=0\)
\(< =>2x^2-2x+5x-5=0\)
\(< =>2x\left(x-1\right)+5\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x+5\right)=0\)
\(< =>\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
\(\hept{\begin{cases}x+2y=1\\-3x+4y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}-3x-6y=-3\\-3x-6y+10y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\10y=-18+3=-15\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x-3=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x=4\\y=-\frac{3}{2}\end{cases}}}}\)
????
xin lỗi nha !
mình mới học lớp 3
mà bài này khó nắm
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
a) Ta có: \(A=\sqrt{12}+2\sqrt{27}-3\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}-12\sqrt{3}\)
\(=-4\sqrt{3}\)
b) Ta có: \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\)
\(=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)
\(=4\sqrt{5a}\)
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)
\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)
\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)
\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)
\(A=\left(4-12+5\right)\sqrt{2}\)
\(A=-3\sqrt{2}\)
b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)
\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)
\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)
\(B=2\sqrt{3}\)
c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)
\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)
\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)
\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)
\(C=\left(2+12-10\right)\sqrt{5a}\)
\(C=4\sqrt{5a}\)
a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\) \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\) b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)