\(\frac{x\left(3x-1\right)}{2-\sqrt{4+x-3x^2}}\le\frac{1}{2}\)
Điều kiện: $ - \frac{1}{3} \le x \le 6$
Ta nhẩm thấy x = 5 là nghiệm của PT, thêm bớt và trục căn thức ta có:
Phương trình $ \Leftrightarrow \left( {\sqrt {3x + 1} - 4} \right) - \left( {\sqrt {6 - x} - 1} \right) + \left( {3{x^2} - 14x - 5} \right) = 0$
$ \Leftrightarrow \frac{{3\left( {x - 5} \right)}}{{\sqrt {3x + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)\left( {x - 5} \right) = 0$
$ \Leftrightarrow \left( {x - 5} \right)\left[ {\frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)} \right] = 0 \Leftrightarrow \left( {x - 5} \right)g\left( x \right) = 0$
Với điều kiện trên ta thấy g(x) > 0 vậy x = 5 là nghiệm của PT.
Bài 1. Tìm điều kiện các BPT sau
a, \(\sqrt{20-x}>\sqrt{3x-6}+1\)
b, \(\frac{\sqrt{9-x^2}}{x-1}>\frac{1}{\sqrt{x}}+1\)
c, \(x+\frac{x+1}{\sqrt{x-4}}>2-\frac{2}{x^2-25}\)
d, \(\sqrt{x}>\sqrt{-x}\)
e, \(3x+\frac{4}{\sqrt{x-5}}\le9+\frac{x}{x-6}\)
f, \(\frac{x+2}{10+3x^2}\ge7+\frac{4}{\left(3x+9\right)^2}\)
g, \(\frac{\sqrt{x+2}}{\sqrt{x-2}}+\frac{1}{\left(x-4\right)\left(x+6\right)}\le\frac{3}{\sqrt{8-x}}\)
h, \(\frac{\sqrt{x+6}}{\left|x\right|-\sqrt{x+6}}\ge\sqrt{16-2x}\)
tìm GTLN
A=\(3x^2\left(8-x^2\right)\) với \(-2\sqrt{2}\le x\le2\sqrt{2}\)
B=4x(8-5x) với \(0\le x\le\frac{8}{5}\)
C=4(x-1)(8-5x) với \(1\le x\le\frac{8}{5}\)
D=x\(\left(3-\sqrt{3}\right)\) với \(0\le x\le\sqrt{3}\)
Tìm GTNN
A=\(\frac{3x}{2}+\frac{2}{x-1}\) với x>1
B=x+\(\frac{2}{3x-1}\) với x>1/3
A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)
\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)
B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)
\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)
\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)
\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)
\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)
\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)
C = \(4\left(x-1\right)\left(8-5x\right)=\frac{4}{5}.\left(5x-5\right)\left(8-5x\right)\)\(\le\frac{4}{5}.\frac{\left(5x-5+8-5x\right)^2}{4}=\frac{9}{5}\)
\(\Rightarrow\)max C = \(\frac{9}{5}\)\(\Leftrightarrow x=\frac{13}{10}\)(thỏa mãn)
D = \(x\left(3-\sqrt{3}\right)\)(quá dễ rồi)
bài 1giải bpt
a) \(\frac{x+2}{3}-x+1>x+3\)
b) \(\frac{3x+5}{2}-1\le\frac{x+2}{3}+x\)
c) \(\frac{\left(x-2\right)\sqrt{x-1}}{\sqrt{x-1}}< 2\)
bài 2 \ giải hệ bpt
a) \(\left\{{}\begin{matrix}2-x>0\\2x+1>x-2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\frac{2x-1}{3}< -x+1\\\frac{4-3x}{2}< 3-x\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}-2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
Mgọi người giúp mình với ạ
\(B=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\\ vs\\ x=\sqrt[3]{2015}\)
Giải các bất phương trình sau:
a) \(\frac{x^2-9x+14}{x^2+9x+14}\ge0\)
b) \(\frac{x^2+1}{x^2+3x-10}< 0\)
c) \(\frac{10-x}{5+x^2}>\frac{1}{2}\)
d) \(\frac{x+1}{x-1}+2>\frac{x-1}{x}\)
e) \(\frac{1}{x+1}+\frac{2}{x+3}\le\frac{3}{x+2}\)
f) \(\frac{x-3}{x+1}-\frac{x-2}{x-1}\le\frac{x^2+4x+15}{x^2-1}\)
g) \(\frac{x^2-4x+3}{x^2-2x}\ge0\)
h) \(\frac{x+2}{3x+1}\le\frac{x-2}{2x-1}\)
i) \(\frac{11x^2-5x+6}{x^2+5x+6}\le x\)
j) \(\frac{\left(1-2x\right)\left(\sqrt{3}x+1\right)}{2\sqrt{2}x-1}\ge0\)
k) \(\frac{\left(5x+1\right)-\left(7x-2\right)}{\left(-x^2-1\right)\left(x^2-4x+4\right)}\le0\)
l) \(\frac{1}{x^2-7x+5}\ge\frac{1}{x^2+2x+5}\)
m) \(\frac{\left(x-1\right)\left(x^3-1\right)}{x^2+\left(1+2\sqrt{2}\right)x+2+\sqrt{2}}\le0\)
Giúp mình hoàn thành các bài tập này với ạ.Cảm ơn rất nhìuuuuuuu @@@
@Akai Haruma
bài 2: giải các bpt sau:
1) (x-2)(\(9-x^2\))≤0
2) (\(x^2-x-6\))(\(x^2-3x+2\))≥0
3) \(\frac{\left(x-2\right)\left(9-x\right)}{x-1}\)≤0
4) \(\frac{x\left(x^2-3x+2\right)}{x+4}\)≥0
5) \(\frac{\left(x+2\right)}{\left(x+1\right)\left(x-2\right)}\)<0
6) \(\frac{\left(x-2\right)\left(9-x^2\right)}{x-1}\)≥0
7) \(\frac{x^2\left(x-3\right)}{3x^2+x-4}\)≥0
8) \(\frac{x^2-3x+2}{9-x}\)≥0
9) \(\frac{x^2+1}{x^2+3x-10}\)≤0
10) \(\frac{x^2-9x+14}{x^2+9x+14}\)≥0
giải các bất phương trình sau:
1) (x-2)(9-x2)≤0
2) (x2-x-6)(x2-3x+2)≥0
3) \(\frac{\left(x-2\right)\left(9-x\right)}{x-1}\)≤0
4) \(\frac{x\left(x^2-3x+2\right)}{x+4}\)≥0
5) \(\frac{\left(x+2\right)}{\left(x+1\right)\left(x-2\right)}\)<0
6) \(\frac{\left(x-2\right)\left(9-x^2\right)}{x-1}\)≥0
7) \(\frac{x^2\left(x-3\right)}{3x^2+x-4}\)≥0
8) \(\frac{x^2-3x+2}{9-x}\)≥0
9) \(\frac{x^2+1}{x^2+3x-10}\)≤0
10) \(\frac{x^2-9x+14}{x^2+9x+14}\)≥0
3.Rút gọn biểu thức :A=
\(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}}\)
mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)
ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)
<=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)
<=> \(A^3=x^3-3x+3A\)
<=> \(A^3-3A-x^3+3x=0\)
<=>\(\left(A^3-x^3\right)-3A+3x=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)
<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)
<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )
vậy \(A=x\)