Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B= \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\)
ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=>\(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\frac{9}{16}\)
=> min B=9/16 kh x=-1/2
C= \(x^2-2xy+y^2+1\)= \(\left(x-y\right)^2+1\)
ta có \(\left(x-y\right)^2\ge0\)=>\(\left(x-y\right)^2+1\ge1\)
=> Min C=1 khi x=y
a) ta có :
\(\Delta'=1^2-\left(-1-m\right)\left(m^2-1\right)=1-\left(-m^2+1-m^3+m\right)=1+m^2-1+m^3-m=m^3+m^2-m=m\left(m^2+m-1\right)\)để phương trình có nghiệm thì \(\Delta\ge0\)
hay \(m\left(m^2+m-1\right)\ge0\)
=> \(\left\{{}\begin{matrix}m\ge0\\m^2+m-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}m\ge0\\\left(m+\dfrac{1}{2}\right)^2\ge\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}m\ge0\\\left[{}\begin{matrix}m+\dfrac{1}{2}\ge\\m+\dfrac{1}{2}\le-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\dfrac{\sqrt{5}}{2}}\)
Bài 1: tính giá trị biểu thức
a) Thay x= 1/2 và y=-1/3 vào biểu thức A, ta được:
A= 3.(1/2)2 .(-1/3)+ 6.(1/2).(-1/3)2+ 3.(1/2).(-1/3)3= -7/8
Vậy giá trị của biểu thức A tại x=1/2 và y=-1/3 là -7/8
a) ta có :(2^14:1024).2^x=128
=>(2^14:2^10).2^x=2^7
=>2^4.2^x=2^7
=>2^x=2^7:2^4
=>2^x=2^3
=>x=3
b) ta có: 3^x+3^x+1+3^x+2=117
=>3^x.(1+3+3^2)=117
=>3^x.13=117
=>3^x=9=3^2
=>x=2
c)ta có 2^x+2^x+1+2^x+2+2^x+3=480
=>2^x.(1+2+2^2+2^3)=480
=>2^x.15=480
=>2^x=480:15=32=2^5
=>x=5
d) ta có: 2^3.32>=2^n>16
=>2^3.2^5>=2^>2^4
=>2^8>=2^n>2^4
=>n=8;7;6;5
còn lại tương tự
h)16^n<32^4
=>(2^4)^n<(2^5)^4
=>2^4n<2^20
=>4n<20
=>n= 0;1;2;3;4
\(2x-49=5.3^2\)
\(\Rightarrow2x-49=5.9\)
\(\Rightarrow2x=45+49\)
\(\Rightarrow2x=94\)
\(\Rightarrow x=94:2\)
\(\Rightarrow x=47\)
Bài 1:
\(P=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)
\(=120\left(3^x+...+3^{x+96}\right)⋮120\)
ĐKXĐ:\(x\ge0;y\ge1;z\ge2\)
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{x+y+z}{2}\)
\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1+2\sqrt{y-1}+1\right)+\left(z-2+2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-2\right)^2=0\)
Mà \(\left\{\begin{matrix}\left(\sqrt{x-1}-1\right)^2\ge0\\\left(\sqrt{y-1}-1\right)^2\ge0\\\left(\sqrt{z-2}-2\right)^2\ge0\end{matrix}\right.\)\(\forall x;y;z\)
\(\Rightarrow\left\{\begin{matrix}\left(\sqrt{x-1}-1\right)^2=0\\\left(\sqrt{y-1}-1\right)^2=0\\\left(\sqrt{z-2}-2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-1}-1=0\\\sqrt{z-2}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x-1=1\\y-1=1\\z-2=4\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x=2\\y=2\\z=6\end{matrix}\right.\)
=> x02 + y02 + z02 = 22 + 22 + 62 = 44
a)(x2-5x+6)(x2-5x+2)-5
Đặt \(x^2-5x+2=t\) ta được:
\(\left(t+4\right)t-16\)\(=t^2+4t-5\)
\(=t^2+5t-t-5\)
\(=t\left(t+5\right)-\left(t+5\right)\)
\(=\left(t-1\right)\left(t+5\right)\)\(=\left(x^2-5x+2-1\right)\left(x^2-5x+2+5\right)\)
\(=\left(x^2-5x+1\right)\left(x^2-5x+7\right)\)
b) (x2+8x-5)(x2+8x+1)-16
Đặt \(t=x^2+8x-5\) ta đc:
\(t\left(t+6\right)-16\)\(=t^2+6t-16\)
\(=t^2+8t-2t-16\)
\(=t\left(t+8\right)-2\left(t+8\right)\)
\(=\left(t-2\right)\left(t+8\right)\)\(=\left(x^2+8x-5-2\right)\left(x^2+8x-5+8\right)\)
\(=\left(x^2+8x-7\right)\left(x^2+8x+3\right)\)
a) \(16x^2-\left(4x-5\right)^2=15\) \(\Leftrightarrow\) \(16x^2-\left(16x^2-40x+25\right)=15\)
\(\Leftrightarrow\) \(16x^2-16x^2+40x-25=15\) \(\Leftrightarrow\) \(40x-25=15\)
\(\Leftrightarrow\) \(40x=40\) \(\Leftrightarrow\) \(x=1\) vậy \(x=1\)
b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow\) \(4x^2+12x+9-4\left(x^2-1\right)=49\)
\(\Leftrightarrow\) \(4x^2+12x+9-4x^2+4=49\)
\(\Leftrightarrow\) \(12x+13=49\) \(\Leftrightarrow\) \(12x=36\) \(\Leftrightarrow\) \(x=\dfrac{36}{12}=3\)vậy \(x=3\)
c) \(\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(\Leftrightarrow\) \(4x^2-1+1-4x+4x^2=18\)\(\Leftrightarrow\) \(8x^2-4x=18\)
\(\Leftrightarrow\) \(8x^2-4x-18=0\)
\(\Delta'=\left(-2\right)^2-8.\left(-18\right)=4+144=148>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{2+\sqrt{148}}{8}=\dfrac{1+\sqrt{37}}{4}\)
\(x_2=\dfrac{2-\sqrt{148}}{8}=\dfrac{1-\sqrt{37}}{4}\)
vậy \(x=\dfrac{1+\sqrt{37}}{4};x=\dfrac{1-\sqrt{37}}{4}\)
Giải:
a) \(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-16x^2-40x+25=15\)
\(\Leftrightarrow-40x+25=15\)
\(\Leftrightarrow-40x=15-25=-10\)
\(\Leftrightarrow x=-\dfrac{10}{-40}=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1^2\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4=49\)
\(\Leftrightarrow12x+9+4=49\)
\(\Leftrightarrow12x=49-9-4\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=\dfrac{36}{12}=3\)
Vậy \(x=3\)
c) \(\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(\Leftrightarrow4x^2-1+1-4x+4x^2=18\)
\(\Leftrightarrow8x^2-4x=18\)
Mình chỉ làm được đến đây thôi, hình như là đề bị sai bạn nhé!
Chúc bạn học tốt!