K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 10 2025
Bài 4:
a:ĐKXĐ: x>=0; x<>1
b: \(A=\frac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\sqrt{x}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
Bài 5:
\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}:\frac{x+16}{\sqrt{x}+2}\)
\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)
Bài 6:
Ta có: \(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{3a-3\sqrt{ab}-2a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}+b}\)
Bài 3:
a: ĐKXĐ: a>0; b>0; a<>b
b: \(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\sqrt{a}-\sqrt{b}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}-\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 10 2025
Bài 4:
a:ĐKXĐ: x>=0; x<>1
b: \(A=\frac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\sqrt{x}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
Bài 5:
\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}:\frac{x+16}{\sqrt{x}+2}\)
\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)
Bài 6:
Ta có: \(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{3a-3\sqrt{ab}-2a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}=\frac{\sqrt{a}-\sqrt{b}}{a+\sqrt{ab}+b}\)
PB
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
18 tháng 8 2025
a: ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)
=>\(AC^2=10^2-6^2=64=8^2\)
=>AC=8(cm)
Xét ΔABC vuông tại A có \(\sin C=\frac{AB}{BC}=\frac{6}{10}=\frac35\)
nên \(\hat{C}\) ≃37 độ
ΔABC vuông tại A
=>\(\hat{B}+\hat{C}=90^0\)
=>\(\hat{B}=90^0-37^0=53^0\)
b: Xét ΔAHB vuông tại H có HE là đường cao
nên \(AE\cdot AB=AH^2\left(1\right)\)
Xét ΔAHC vuông tại H có HF là đường cao
nên \(AF\cdot AC=AH^2\) (2)
Từ (1),(2) suy ra \(AE\cdot AB=AF\cdot AC\)
c: Xét ΔABC vuông tại A có AH là đường cao
nên \(AH\cdot BC=AB\cdot AC\)
\(AE\cdot AB=AH^2\)
=>\(AE=\frac{AH^2}{AB}\)
\(AF\cdot AC=AH^2\)
=>\(AF=\frac{AH^2}{AC}\)
Xét tứ giác AEHF có \(\hat{AEH}=\hat{AFH}=\hat{FAE}=90^0\)
nên AEHF là hình chữ nhật
=>\(S_{AEHF}=AE\cdot AF=\frac{AH^2}{AB}\cdot\frac{AH^2}{AC}=\frac{AH^4}{AH\cdot BC}=\frac{AH^3}{BC}\)
VT
0
NH
0
NT
0
f: \(F=\sqrt[3]{4+\frac53\cdot\sqrt{\frac{31}{3}}}+\sqrt[3]{4-\frac53\cdot\sqrt{\frac{31}{3}}}\)
=>\(F^3=4+\frac53\cdot\sqrt{\frac{31}{3}}+4-\frac53\cdot\sqrt{\frac{31}{3}}+3\cdot F\cdot\sqrt[3]{\left(4+\frac53\sqrt{\frac{31}{3}}\right)\left(4-\frac53\cdot\sqrt{\frac{31}{3}}\right)}\)
=>\(F^3=8+3\cdot F\cdot\sqrt[3]{16-\frac{25}{9}\cdot\frac{31}{3}}=8+3F\cdot\sqrt[3]{-\frac{343}{27}}=8-7F\)
=>\(F^3+7F-8=0\)
=>\(F^3-F^2+F^2-F+8F-8=0\)
=>\(\left(F-1\right)\left(F^2+F+8\right)=0\)
=>F-1=0
=>F=1
o: \(\sqrt[3]{6\sqrt3-10}=\sqrt[3]{3\sqrt3-3\cdot\left(\sqrt3\right)^2\cdot1+3\cdot\sqrt3\cdot1^2-1}=\sqrt[3]{\left(\sqrt3-1\right)^3}=\sqrt3-1\)
\(\sqrt{2+\sqrt3}=\frac{\sqrt{4+2\sqrt3}}{\sqrt2}=\frac{\sqrt{\left(\sqrt3+1\right)^2}}{\sqrt2}=\frac{\sqrt3+1}{\sqrt2}\)
\(X=\sqrt{2+\sqrt3}-\sqrt{\frac{\sqrt[3]{6\sqrt3-10}}{\sqrt3+1}}=\frac{\sqrt3+1}{\sqrt2}-\sqrt{\frac{\sqrt3-1}{\sqrt3+1}}\)
\(=\frac{\sqrt3+1}{\sqrt[2]{2}}-\sqrt{\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^2}}=\frac{\sqrt3+1}{\sqrt2}-\sqrt{\frac{2}{\left(\sqrt3-1\right)^2}}\)
\(=\frac{\sqrt3+1}{\sqrt2}-\frac{\sqrt2}{\sqrt3+1}=\frac{\left(\sqrt3+1\right)^2-2}{\sqrt2\left(\sqrt3+1\right)}=\frac{4+2\sqrt3-2}{\sqrt2\left(\sqrt3+1\right)}=\frac{2\sqrt3+2}{\sqrt2\left(\sqrt3+1\right)}\)
\(=\frac{2\left(\sqrt3+1\right)}{\sqrt2\left(\sqrt3+1\right)}=\frac{2}{\sqrt2}=\sqrt2\)
p: \(K=\frac{\left(\sqrt3-1\right)\cdot\sqrt[3]{10+6\sqrt3}}{\sqrt{21+4\sqrt5}+3}=\frac{\left(\sqrt3-1\right)\cdot\sqrt[3]{3\sqrt3+3\cdot\left(\sqrt3\right)^2\cdot1+3\cdot\sqrt3\cdot1^2+1^3}}{\left(\sqrt{\left(2\sqrt5+1\right)^2}\right)+3}\)
\(=\frac{\left(\sqrt3-1\right)\cdot\sqrt[3]{\left(\sqrt3+1\right)^2}}{2\sqrt5+1+3}=\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}{2\sqrt5+4}=\frac{2}{2\sqrt5+4}=\frac{1}{\sqrt5+2}=\sqrt5-2\)
q: \(Q=\sqrt[3]{55+\sqrt{3024}}+\sqrt[3]{55-\sqrt{3024}}\)
=>\(Q^3=55+\sqrt{3024}+55-\sqrt{3024}+3\cdot Q\cdot\sqrt[3]{\left(55-\sqrt{3024}\right)\left(55+\sqrt{3024}\right)}\)
=>\(Q^3=110+3\cdot Q\cdot\sqrt[3]{3025-3024}=110+3Q\)
=>\(Q^3-3Q-110=0\)
=>\(Q^3-5Q^2+5Q^2-25Q+22Q-110=0\)
=>\(\left(Q-5\right)\left(Q^2+5\cdot Q+22\right)=0\)
=>Q-5=0
=>Q=5