tìm giá trị lớn nhất các biểu thức sau :
a,A= căn (x-2) + căn (4+x)
b, B= căn (6-x) + căn (x+2)
c, C= căn x + căn...
Đọc tiếp
tìm giá trị lớn nhất các biểu thức sau :
a,A= căn (x-2) + căn (4+x)
b, B= căn (6-x) + căn (x+2)
c, C= căn x + căn (2-x)
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BN
2
MH
23 tháng 8 2022
Điều kiện: \(x\ge1\)
\(\sqrt{x+2\sqrt{x-1}}=2\\ \Rightarrow\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\\ \Rightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\\ \Rightarrow\sqrt{x-1}+1=2\\ \Rightarrow\sqrt{x-1}=1\\ \Rightarrow x-1=1\\ \Rightarrow x=2\left(TM\right)\)
24 tháng 8 2022
√x+2√x−1=2x+2x−1=2⇔√(x−1)+2.√x−1.1+1=2⇔(x−1)+2.x−1.1+1=2
⇔√(√x−1+1)2=2⇔(x−1+1)2=2
⇔√x−1+1=2⇔x−1+1=2
⇔⇔√x−1=1x−1=1
⇔(√
BN
2
BN
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
29 tháng 12 2025
\(x^2-7\)
\(=x^2-\left(\sqrt7\right)^2\)
\(=\left(x-\sqrt7\right)\left(x+\sqrt7\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 12 2025
a: \(A=\frac{5+\sqrt5}{\sqrt5+2}+\frac{\sqrt5}{\sqrt5-1}-\frac{3\sqrt5}{3+\sqrt5}\)
\(=\frac{\left(5+\sqrt5\right)\left(\sqrt5-2\right)}{\left(\sqrt5+2\right)\left(\sqrt5-2\right)}+\frac{\sqrt5\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}-\frac{3\sqrt5\left(3-\sqrt5\right)}{\left(3+\sqrt5\right)\left(3-\sqrt5\right)}\)
\(=\left(5+\sqrt5\right)\left(\sqrt5-2\right)+\frac{\sqrt5\left(\sqrt5+1\right)}{4}-\frac{3\sqrt5\left(3-\sqrt5\right)}{4}\)
\(=5\sqrt5-10+5-2\sqrt5+\frac{5+\sqrt5-9\sqrt5+15}{4}\)
\(=3\sqrt5-5-2\sqrt5+5=\sqrt5\)
b: Đặt \(C=\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2\)
\(=2+\sqrt3+3-\sqrt5+2\cdot\sqrt{\left(2+\sqrt3\right)\left(3-\sqrt5\right)}\)
\(=5+\sqrt3-\sqrt5+\sqrt{\left(4+2\sqrt3\right)\left(6-2\sqrt5\right)}\)
\(=5+\sqrt3-\sqrt5+\sqrt{\left(\sqrt3+1\right)^2}\cdot\sqrt{\left(\sqrt5-1\right)^2}\)
\(=5+\sqrt3-\sqrt5+\left(\sqrt3+1\right)\left(\sqrt5-1\right)\)
\(=5+\sqrt3-\sqrt5+\sqrt{15}-\sqrt3+\sqrt5-1=\sqrt{15}+4\)
Đặt \(D=\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2\)
\(=2-\sqrt3+3+\sqrt5+2\cdot\sqrt{\left(2-\sqrt3\right)\left(3+\sqrt5\right)}\)
\(=5-\sqrt3+\sqrt5+\sqrt{\left(4-2\sqrt3\right)\left(6+2\sqrt5\right)}\)
\(=5-\sqrt3+\sqrt5+\sqrt{\left(\sqrt3-1\right)^2}\cdot\sqrt{\left(\sqrt5+1\right)^2}\)
\(=5-\sqrt3+\sqrt5+\left(\sqrt3-1\right)\left(\sqrt5+1\right)=5-\sqrt3+\sqrt5+\sqrt{15}+\sqrt3-\sqrt5-1=4+\sqrt{15}\)
Ta có: \(B=21\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2-6\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2-15\sqrt{15}\)
\(=21\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=15\left(4+\sqrt{15}\right)-15\sqrt{15}=60\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
21 tháng 12 2025
a: \(A=\frac{5+\sqrt5}{\sqrt5+2}+\frac{\sqrt5}{\sqrt5-1}-\frac{3\sqrt5}{3+\sqrt5}\)
\(=\frac{\left(5+\sqrt5\right)\left(\sqrt5-2\right)}{\left(\sqrt5+2\right)\left(\sqrt5-2\right)}+\frac{\sqrt5\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}-\frac{3\sqrt5\left(3-\sqrt5\right)}{\left(3+\sqrt5\right)\left(3-\sqrt5\right)}\)
\(=\left(5+\sqrt5\right)\left(\sqrt5-2\right)+\frac{\sqrt5\left(\sqrt5+1\right)}{4}-\frac{3\sqrt5\left(3-\sqrt5\right)}{4}\)
\(=5\sqrt5-10+5-2\sqrt5+\frac{5+\sqrt5-9\sqrt5+15}{4}\)
\(=3\sqrt5-5-2\sqrt5+5=\sqrt5\)
b: Đặt \(C=\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2\)
\(=2+\sqrt3+3-\sqrt5+2\cdot\sqrt{\left(2+\sqrt3\right)\left(3-\sqrt5\right)}\)
\(=5+\sqrt3-\sqrt5+\sqrt{\left(4+2\sqrt3\right)\left(6-2\sqrt5\right)}\)
\(=5+\sqrt3-\sqrt5+\sqrt{\left(\sqrt3+1\right)^2}\cdot\sqrt{\left(\sqrt5-1\right)^2}\)
\(=5+\sqrt3-\sqrt5+\left(\sqrt3+1\right)\left(\sqrt5-1\right)\)
\(=5+\sqrt3-\sqrt5+\sqrt{15}-\sqrt3+\sqrt5-1=\sqrt{15}+4\)
Đặt \(D=\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2\)
\(=2-\sqrt3+3+\sqrt5+2\cdot\sqrt{\left(2-\sqrt3\right)\left(3+\sqrt5\right)}\)
\(=5-\sqrt3+\sqrt5+\sqrt{\left(4-2\sqrt3\right)\left(6+2\sqrt5\right)}\)
\(=5-\sqrt3+\sqrt5+\sqrt{\left(\sqrt3-1\right)^2}\cdot\sqrt{\left(\sqrt5+1\right)^2}\)
\(=5-\sqrt3+\sqrt5+\left(\sqrt3-1\right)\left(\sqrt5+1\right)=5-\sqrt3+\sqrt5+\sqrt{15}+\sqrt3-\sqrt5-1=4+\sqrt{15}\)
Ta có: \(B=21\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2-6\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2-15\sqrt{15}\)
\(=21\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=15\left(4+\sqrt{15}\right)-15\sqrt{15}=60\)
BN
1
23 tháng 8 2022
Ta có: \(m_{H_2SO_4\left(TT\right)}=5\left(tấn\right)\)
Mà `H = 75\%`
=> \(m_{H_2SO_4\left(LT\right)}=\dfrac{5}{75\%}=\dfrac{20}{3}\left(tấn\right)=\dfrac{20000}{3}\left(kg\right)\)
=> \(n_{H_2SO_4\left(LT\right)}=\dfrac{20000}{3.98}=\dfrac{10000}{147}\left(kmol\right)\)
BTNT S: \(n_{FeS_2}=\dfrac{1}{2}n_{H_2SO_4}=\dfrac{5000}{147}\left(kmol\right)\)
=> \(m_{quặng}=\dfrac{\dfrac{5000}{147}.120}{60\%}=6802,72\left(kg\right)\)
23 tháng 8 2022
\(m_{H_2SO_4\left(TT\right)}=4.85\%=3,4\left(tấn\right)=3400\left(kg\right)\)
=> \(m_{H_2SO_4\left(LT\right)}=\dfrac{3,4}{80\%}=4250\left(kg\right)\)
=> \(n_S=n_{H_2SO_4}=\dfrac{4250}{98}=\dfrac{2125}{49}\left(kmol\right)\)
=> \(m_{quặng}=\dfrac{\dfrac{2125}{49}.32}{32\%}=\dfrac{212500}{49}\left(kg\right)\)