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ĐKXĐ: \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2+\sqrt{x}-2-2\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{x-4}\)
\(A=10-\left(2\sqrt{2}-3\sqrt{3}\right)\left(-2\sqrt{2}-3\sqrt{3}\right)\)
\(=10+\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)\)
\(=10+\left(8-27\right)=-9\)
\(AB=-1\Leftrightarrow\frac{-9}{x-4}=-1\Rightarrow x-4=9\Rightarrow x=13\)
1)
dat \(a=\sqrt[3]{x+1};b=\sqrt[3]{7-x}\)
ta co b=2-a
a^3+b^3=x+1+7-x=8
a^3+b^3=a^3+b^3+3ab(a+b)
ab(a+b)=0
suy ra a=0 hoac b=0 hoac a=-b
<=> x=-1; x=7
a=-b
a^3=-b^3
x+1=x+7 (vo li nen vo nghiem)
cau B tuong tu
2)
tat ca cac bai tap deu chung 1 dang do la
\(\sqrt[3]{a+m}+\sqrt[3]{b-m}\)voi m la tham so
dang nay co 2 cach
C1 lap phuong VD: \(B^3=10+3\sqrt[3]{< 5+2\sqrt{13}>< 5-2\sqrt{13}>}\left(B\right)\)
B^3=10-9B
B=1 cach nay nhanh nhung kho nhin
C2 dat an
\(a=\sqrt[3]{5+2\sqrt{13}};b=\sqrt[3]{5-2\sqrt{13}}\)
de thay B=a+b
a^3+b^3=10
ab=-3
B^3=10-9B
suy ra B=1
tuong tu giai cac cau con lai.
Bài 1:
a. Đặt \(a=\sqrt[3]{x+1}\); \(b=\sqrt[3]{7-x}\). Ta có:
\(\hept{\begin{cases}a+b=2\\a^3+b^3=8\end{cases}\Leftrightarrow a^3+\left(2-a\right)^3=8\Leftrightarrow...\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}a=0\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=0\\\sqrt[3]{7-x}=2\end{cases}}\)hoặc \(\hept{\begin{cases}\sqrt[3]{x+1}=2\\\sqrt[3]{7-x}=0\end{cases}}\)
\(\Leftrightarrow x=-1\)hoặc \(x=7\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{2}{\sqrt{x}-1}\)
\(=\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\frac{2}{\sqrt{x}-1}\right)\)
\(=\frac{\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\left(\sqrt{x}-1\right)}=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{2}{x+\sqrt{x}+1}\)
Ta có \(x\ge0\Rightarrow x+\sqrt{x}\ge0\Rightarrow x+\sqrt{x}+1>0\)
\(\Rightarrow A=\frac{2}{x+\sqrt{x}+1}>0\)
Mặt khác cũng do \(x+\sqrt{x}+1\ge1\Rightarrow P\le\frac{2}{1}=2\)
\(\Rightarrow A_{max}=2\) khi \(x=0\)
Lời giải:
ĐK: $x\geq 0$
a)
\(A=\frac{\sqrt{x}(\sqrt{x^3}+1)}{x-\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}\)
\(=\sqrt{x}(\sqrt{x}+1)-\sqrt{x}(\sqrt{x}-1)=2\sqrt{x}\)
b)
\(x=29-12\sqrt{5}=20+9-2\sqrt{20.9}=(\sqrt{20}-\sqrt{9})^2\Rightarrow \sqrt{x}=\sqrt{20}-3\)
Do đó: $A=2\sqrt{x}=2(\sqrt{20}-3)$
c)
$x+A=m\Leftrightarrow x+2\sqrt{x}=m$ (đề không đủ ý)
\(a,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow A=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow A=\frac{x+2+x-\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow\frac{3x+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(b,Tacó:P=\frac{A}{B}=\frac{3x+3}{2\left(x+\sqrt{x}+1\right)}\)
\(\Rightarrow P=\frac{3}{2}.\frac{x+1}{x+\sqrt{x}+1}\)
\(\Rightarrow P=\frac{3}{2}.\frac{x+1}{x+1+\sqrt{x}}\)
\(\Rightarrow P=\frac{3}{2}.\left(1-\frac{\sqrt{x}}{x+1+\sqrt{x}}\right)\)
\(\Rightarrow P\le\frac{3}{2}.\left(1-0\right)\)
\(\Rightarrow P\le\frac{3}{2}\)
\(\Rightarrow Max_P=\frac{3}{2}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{-x+x\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x-\sqrt{x}-x+x\sqrt{x}+6-x-\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\sqrt{x}-x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(x-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\sqrt{x}-2\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(Q=\frac{\left(x+27\right)P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{x+27}{\sqrt{x}+3}\)
\(Q=\frac{x+27}{\sqrt{x}+3}\ge6\\ \Leftrightarrow\frac{x+27}{\sqrt{x}+3}-6\ge0\\ \Leftrightarrow\frac{x+27-6\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{x-6\sqrt{x}+45}{\sqrt{x}+3}\ge0\)
Dễ thấy \(x-6\sqrt{x}+45=\left(\sqrt{x}-3\right)^2+36\ge36>0\forall x\ge0\)
\(\sqrt{x}+3\ge3>0\forall x\ge0\)
=> Ko có giá trị nào của x thỏa mãn yêu cầu
P/s: Nếu đề là \(x\sqrt{x}+27\)thì sẽ khác một chút :v
Bạn ơi chỗ kia phải là \(\frac{x-6\sqrt{x}+9}{\sqrt{x}+3}\)
\(M=4x^2-2\left(a+b+c\right)x-\left(ab+bc+ca\right)\)
Thay x, ta có:
\(M=4.\left(\frac{a+b+c}{2}\right)^2-2\left(a+b+c\right).\frac{a+b+c}{2}-\left(ab+bc+ca\right)\)
\(=\left(a+b+c\right)^2-\left(a+b+c\right)^2-\left(ab+bc+ca\right)\)
\(=-ab-bc-ca\)
2/ Số mũ tùm lum, có lẽ b nên ktra lại đề bài!