Tính \(A=\sqrt{8-a}+\sqrt{5+a}\) biết \(\sqrt{\left(8-a\right)\left(5+a\right)}=34\)
Giá trị biểu thức \(\sqrt{8-a}+\sqrt{a+5}\)biết \(\sqrt{\left(8-a\right)\left(5+a\right)}=34\)
\(\left(\sqrt{8-a}+\sqrt{a+5}\right)^2=8-a+a+5+2.\sqrt{\left(8-a\right)\left(a+5\right)}=13+2.34=13+68=81\)
=>\(\sqrt{8-a}+\sqrt{a+5}=\sqrt{81}=9\)
Ko biết đâu mà hỏi ,tôi mới có học lớp 6 thôi
\(A=\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3}-\sqrt{5}\)
B=\(B=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8}-2\sqrt{15}\)
\(Cho\sqrt{8-a}+\sqrt{5+a}=5tinh\sqrt{\left(8-a\right)\left(5+a\right)}\)
a: \(A=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8\)
b: \(B=\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\left(5-3\right)=2\cdot2=4\)
cho \(\sqrt{8-a}+\sqrt{5+a}\)=5
tính \(\sqrt{\left(8-a\right)\left(5+a\right)}\)
\(\sqrt{8-a}+\sqrt{5+a}=5\left(Đk:-5\le a\le8\right)\)
\(8-a+5+a+2\sqrt{\left(8-a\right)\left(5+a\right)}=25\)
\(\sqrt{\left(8-a\right)\left(5+a\right)}=6\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
rút gọn
a) \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
b) \(\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
c) \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
Với $a,b,c$ là các số thực thỏa mãn $a+b+c=3.$ Chứng minh rằng$:$
$$ \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} + \frac 1{8 + 5(a^2 + b^2)}\le \frac 1{6},$$
Đẳng thức xảy ra khi và chỉ khi $a=b=c$ hoặc $a=b=\frac{c}{13}$ và các hoán vị.
Ji Chen đưa ra một kiểu SOS rất khủng$:$
$\frac{11664}{5}\left(\frac{1}{6}-\sum{\frac{1}{8+5b^2+5c^2}}\right)\prod{\left(8+5b^2+5c^2\right)}$
$=23050\prod{(b-c)^2}+\sum{\Big\{\left(a^2-bc\right)\left[5\left(66-7\sqrt{34}\right)-25\left(2-3\sqrt{34}\right)a\right]}$
${-(a-1)\left[75\sqrt{34}a^2+15\left(18+5\sqrt{34}\right)a+6\left(23-5\sqrt{34}\right)\right]\Big\}^2}\geq 0.$
Mặt khác bài này rất hay và có nhiều kiểu SOS đẹp, mọi người thử tìm xem$?$
Hay thậm chí là những cách giải khác ngoài SOS cho bài này$?$
Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
Thực hiện các phép tính (không được ghi mỗi kết quả không, phải giải chi tiết)
A = \(2\sqrt{10}.3\sqrt{8}.2\)
B = \(\sqrt{20}\left(2\sqrt{3}-\sqrt{5}\right)\)
C = \(\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)\)
a: \(=12\sqrt{80}=48\sqrt{5}\)
b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)
c: =20-9=11
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)