giúp mình với mn
giải bất pt:
a.\(\left\{{}\begin{matrix}2+x< 3+3x\\x-2\ge1-3x\end{matrix}\right.\)
b.\(\frac{1}{x-1}\le x-1\)
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3+3x=448y^3+6y\\385x^2-16y^2=96\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}3x^3-y^3=\dfrac{1}{x+y}\\x^2+y^2=1\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-448y^3=-3x+6y\\96=385x^2-16y^2\end{matrix}\right.\)
\(\Rightarrow96\left(x^3-448y^3\right)=\left(-3x+6y\right)\left(385x^2-16y^2\right)\)
\(\Leftrightarrow\left(x-4y\right)\left(417x^2+898xy+3576y^2\right)=0\)
\(\Leftrightarrow x-4y=0\)
\(\Leftrightarrow x=4y\)
Thế vào \(385x^2-16y^2=96\)
\(\Rightarrow...\)
b.
ĐKXĐ: \(x+y\ne0\)
\(\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=x^2+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=\left(x^2+y^2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left(3x^3-y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)\left(2x^2+xy+y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
Thế vào \(x^2+y^2=1\)...
giải các hệ BPT sau:
a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)
GIẢI CÁC HPT SAU:
a) \(\left\{{}\begin{matrix}3x-5\ge2x+12\\\frac{x-3}{2}\le\frac{2x+27}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2-7x\ge x-14\\3x+1\ge6x-11\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x+\frac{3}{5}< x+2\\2x+3 >\frac{7x-3}{4}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x+5}{2}< 4x-3\\x\left(x-1\right)\ge\left(x-3\right)\left(4+x\right)\end{matrix}\right.\)
GIÚP MIK VỚI
Tìm m để hệ bất phương trình vô nghiệm
a) \(\left\{{}\begin{matrix}3x+4>x+9\\1-2x\le m-3x+1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+7\ge8x+1\\m+5< 2x\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge x^2+7x+1\\2m\le8+5x\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}3x+5\ge x-1\\\left(x+2\right)^2\le\left(x-1\right)^2+9\\mx+1>\left(m-2\right)x+m\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}2\left(x-3\right)< 5\left(x-4\right)\\mx+1\le x-1\end{matrix}\right.\)
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
Tập nghiệm của bất phương trình (x2-3x)\(\sqrt{2x^2-3x-2}\ge0\) Là
A.\(\left[{}\begin{matrix}X\ge3\\x=2\\x\le\frac{-1}{2}\end{matrix}\right.\)
B.\(\left[{}\begin{matrix}X\ge3\\x\le0\end{matrix}\right.\)
C.\(\left[{}\begin{matrix}X\ge2\\x\le\frac{-1}{2}\end{matrix}\right.\)
D.x∈\(\left\{\frac{-1}{2};0;2;3\right\}\)
Mình đã giải đã có ở đây:
Câu hỏi của Nguyễn Vi - Toán lớp 10 | Học trực tuyến
Đáp án A.
1, Giải bất pt sau:
\(-2x+\frac{3}{5}\le\frac{3\left(2x-7\right)}{3}\)
2, Xác định m để hệ bất pt sau có nghiệm:
a, \(\left\{\begin{matrix}x+m-1>0\\3m-2-x>0\end{matrix}\right.\)
b, \(\left\{\begin{matrix}x-1>0\\mx-3>0\end{matrix}\right.\)
Bai1:
\(-2x+\frac{3}{5}\le\frac{3\left(2x-7\right)}{3}\Leftrightarrow-10x+3\le5\left(2x-7\right)\Leftrightarrow-10x+3\le10x-35\)
\(\Leftrightarrow\left(10+10\right)x\ge3+35\Rightarrow x\ge\frac{38}{20}=\frac{19}{10}\)
Bài
\(\left\{\begin{matrix}x+m-1>0\\3m-2-x>0\end{matrix}\right.\Leftrightarrow\left(I\right)\left\{\begin{matrix}x>1-m\\x< 3m-2\end{matrix}\right.\)
Hệ (I) có nghiệm cần m thỏa mãn:
\(1-m< 3m-2\Leftrightarrow1+2< 3m+m\Rightarrow m>\frac{3}{2}\)
Kết luận: để hệ có nghiệm cần: m>3/2
Tìm m để hệ bất phương trình có nghiệm duy nhất
a) \(\left\{{}\begin{matrix}2x-1\ge3\\x-m\le0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m^2x\ge6-x\\3x-1\le x+5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge x^2+7x+1\\2m\le8+5x\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}mx\le m-3\\\left(m+3\right)x\ge m-9\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}2m\left(x+1\right)\ge x+3\\4mx+3\ge4x\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le m\end{matrix}\right.\)
Hệ có nghiệm duy nhất \(\Leftrightarrow m=2\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x\ge6\\2x\le6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{6}{m^2+1}\\x\le3\end{matrix}\right.\)
Hệ có nghiệm duy nhất \(\Leftrightarrow\dfrac{6}{m^2+1}=3\)
\(\Leftrightarrow m=\pm1\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-6x+9\ge x^2+7x+1\\5x\ge2m-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{8}{13}\\x\ge\dfrac{2m-8}{5}\end{matrix}\right.\)
Pt có nghiệm duy nhất khi \(\dfrac{2m-8}{5}=\dfrac{8}{13}\Leftrightarrow m=\dfrac{72}{13}\)
d.
Hệ có nghiệm duy nhất khi:
TH1:
\(\left\{{}\begin{matrix}m>0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-9=m^2-9m\end{matrix}\right.\) \(\Leftrightarrow m=1\)
TH2:
\(\left\{{}\begin{matrix}m+3< 0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)
\(\Leftrightarrow m=1\) (ktm)
Vậy \(m=1\)
e.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m-1\right)x\ge-2m+3\\\left(4-4m\right)x\le3\end{matrix}\right.\)
Hệ có nghiệm duy nhất khi:
\(\left\{{}\begin{matrix}\left(2m-1\right)\left(4-4m\right)>0\\\dfrac{-2m+3}{2m-1}=\dfrac{3}{4-4m}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}< m< 1\\\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow m=\dfrac{3}{4}\)
Giải các hệ bất phương trình:
a) \(\left\{{}\begin{matrix}4x^2-5x-6\le0\\\left(1-x^2\right)\left(4x^2-12x+5\right)>0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-x-2\ge0\\2x^2-11x+9< 0\\x^3-x^2+2x-2>0\end{matrix}\right.\)
c) \(-3\le\frac{x^2-3x-1}{x^2+x+1}< 3\)