1, So sánh
(-0,25)9 và (-0,05)7
2. Tìm x
a) x - \(\left|x-3\right|\)= 3
b) 7 - \(\left|x+3\right|\)< 2
c) \(\left|3x-2\right|\) = x+1
d) \(\left|x-3\right|\) = 7-x
Lưu ý; Từ baì 2c,d ; nhớ ra điều kiện trước khi làm
Làm giúp mik với, thanks
tìm khoảng đồng biến và nghịch biến của hàm số sau
a) \(y'=\left(x-3\right)^3\left(x-1\right)^{22}\left(-3x-6\right)^7\)
b) \(y'=\left(4x-3\right)^3\left(x^2-1\right)^{21}\left(3x-9\right)^7\)
a: \(y'< 0\)
=>\(\left(x-3\right)^3\cdot\left(x-1\right)^{22}\cdot\left(-3x-6\right)^7< 0\)
=>\(\left(x-3\right)\left(-3x-6\right)< 0\)
=>\(\left(x+2\right)\left(x-3\right)>0\)
=>\(\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
y'>0
=>\(\left(x+2\right)\left(x-3\right)< 0\)
=>\(-2< x< 3\)
y'=0
=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\\-3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)
Ta có bảng xét dấu sau:
x | \(-\infty\) -2 1 3 +\(\infty\) |
y' | - 0 + 0 + 0 - |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-2;1\right);\left(1;3\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-\infty;-2\right);\left(3;+\infty\right)\)
b: y'<0
=>\(\left(4x-3\right)^3\cdot\left(x^2-1\right)^{21}\left(3x-9\right)^7< 0\)
=>\(\left(4x-3\right)\left(3x-9\right)\left(x^2-1\right)< 0\)
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{3}{4}\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)
y'>0
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\-1< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{3}{4}< x< 1\)
Ta sẽ có bảng xét dấu sau đây:
x | \(-\infty\) -1 3/4 1 3 +\(\infty\) |
y' | + 0 - 0 + 0 - 0 + |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;-1\right);\left(\dfrac{3}{4};1\right);\left(3;+\infty\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-1;\dfrac{3}{4}\right);\left(1;3\right)\)
1.Cho \(r\left(x\right)=-\left(3x-7\right)^2+2\left(3x-7\right)-17\)
Tìm GTLN của biểu thức r(x).
2. So sánh : \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)với \(B=3^{32}-1\)
3. Tìm x, y biết: \(y^2+2y+4x-2^{x+1}+2=0\)
Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
1, Khai triển ra ta được:
\(r\left(x\right)=-\left(9x^2-42x+49\right)+6x-14-17\)
\(=-9x^2+42x-49+6x-14-17\)
\(=-9x^2+48x-80\)
\(=-9x^2+48x-64-16\)
\(=-\left(\left(3x\right)^2-3x.2.8+8^2\right)-16\)
\(=-\left(3x+8\right)^2-16\)
\(Do-\left(3x+8\right)^2\le0\)
\(=>-\left(3x+8\right)^2-16\le-16\)
Dấu bằng xảy ra khi \(3x+8=0=>x=-\frac{8}{3}\)
Vậy giá trị nhỏ nhất là -16 tại \(x=-\frac{8}{3}\)
Đạo hàm của hàm số \(y=\left(-x^2+3x+7\right)^7\) là:
A. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
B. \(y'=7\left(-x^2+3x+7\right)^6\)
C. \(y'=\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
D. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
Tìm x, biết:
\(a,\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-1}{6}\)
\(b,\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(c,\dfrac{x}{8}=\dfrac{9}{4}\)
\(d,\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(e,4,5x-6,2x=6,12\)
\(h,11,4-\left(x-3,4\right)=-16,2\)
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
Giair phương trình sau:
a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)
c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)
g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)
i,\(x^6-x^2=0\) j,\(x^3-12=13x\)
k,\(-x^5+4x^4=-12x^3\) l, \(x^3=4x\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)