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Bài 1:
\(\left(2x+1\right)^3=9\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)^3-9\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[\left(2x+1\right)^2-9\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)\left(2x+1+3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-2\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=1\\x=-2\end{array}\right.\)
Bài 2:
\(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\)
Vì: \(\left(2x-1\right)^2+\left(3-y\right)^2\ge0\)
=> \(\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)
Dấu "=" xảy ra khi \(x=\frac{1}{2};y=3\)
Vậy GTNN của A là 2017 khi \(x=\frac{1}{2};y=3\)
Bài 1:
(2x + 1)3 = 9.(2x + 1)
=> (2x + 1)3 - 9.(2x + 1) = 0
=> (2x + 1).[(2x + 1)2 - 9] = 0
=> (2x + 1).(2x + 1 - 3).(2x + 1 + 3) = 0
=> (2x + 1).(2x - 2).(2x + 4) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=-1\\2x=2\\2x=-4\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-1}{2}\\x=1\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{\frac{-1}{2};1;-2\right\}\)
Bài 2:
Có: \(\left(2x-1\right)^2\ge0;\left(3-y\right)^2\ge0\forall x;y\)
=> \(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)
Dấu "=" xảy ra khi và chỉ khi \(\begin{cases}\left(2x-1\right)^2=0\\\left(3-y\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}2x-1=0\\3-y=0\end{cases}\)\(\Rightarrow\begin{cases}2x=1\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=3\end{cases}\)
Vậy GTNN của A là 2017 khi và chỉ khi \(x=\frac{1}{2};y=3\)
Với mọi x thuộc R Có (x^2-9)^2 \(\ge\) 0
[y-4] \(\ge\) 0
Suy ra (x^2-9)^2+[y-4] - 1 \(\ge\) -1
Xét A=-1 khi và chỉ khi (x^2-9)^2 và [y-4] đều bằng 0
Tự tính ra
Xin lỗi nhưng vì không biết nên mình phải dùng [ ] thay cho GTTĐ nhé
Xin lỗi nhiều tại mình o tìm được kí hiệu đó
GIẢI CHI TIẾT RA GIÚP MK VS NHÉ!!!!MK RẤT CẢM ƠN
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
\(\dfrac{3}{7}-x=\dfrac{1}{4}-\left(-\dfrac{3}{5}\right)\)
\(\Rightarrow\dfrac{3}{7}-x=\dfrac{17}{20}\)
\(\Rightarrow x=\dfrac{-59}{140}\)
Vậy \(x=\dfrac{-59}{140}.\)
Lần sau tự làm mấy bài này đi bạn
\(\dfrac{-3}{26}+2\dfrac{4}{69}=\dfrac{-3}{26}+2+\dfrac{4}{69}=\left(\dfrac{-3}{26}+\dfrac{4}{69}\right)+2=\dfrac{-103}{1794}+2=1,9425...\)
Máy mk ko quy đổi được về phân số bạn thông cảm trần thị anh thư
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
Ta có:A=\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\frac{1}{2}\)\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\frac{1}{2}A-A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)-\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(-\frac{1}{2}A\)=\(\left(\frac{1}{2}^{100}\right)-\frac{1}{2}\)
\(-\frac{1}{2}A\)=\(-\frac{1}{2}\)
A=\(-\frac{1}{2}:\left(-\frac{1}{2}\right)\)
A=1
Chúc bạn học tốt!
F=|x-1|+|x-2|+|x-3|+...+|x-100|=|x-1|+|2-x|+|x-3|+...+|100-x|
Áp dụng bđt |a|+|b|\(\ge\)|a+b|, ta có:
F=|x-1|+|2-x|+|x-3|+...+|100-x| \(\ge\) |x-1+2-x+x-3+...+100-x| = |50| = 50
=> F\(\ge\)50 => \(Min_F=50\)
P/s: mấy thánh toán đi ngang cho mik hỏi giải vậy có đúng hog?
\(F=\left|x-1\right|+\left|x-2\right|+....+\left|x-99\right|+\left|x-100\right|\)
\(F=\left(\left|x-1\right|+\left|x-100\right|\right)+\left(\left|x-2\right|+\left|x-99\right|\right)+.....+\left(\left|x-50\right|+\left|x-51\right|\right)\)
\(F=\left(\left|x-1\right|+\left|100-x\right|\right)+\left(\left|x-2\right|+\left|99-x\right|\right)+....+\left(\left|x-50\right|+\left|51-x\right|\right)\)
(do \(\left|-A\left(x\right)\right|=\left|A\left(x\right)\right|\))
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x-1\right|\ge1;\left|x-2\right|\ge x-2;.....;\left|99-x\right|\ge99-x;\left|100-x\right|\ge100-x\)
\(\Rightarrow\left|x-1\right|+\left|100-x\right|\ge x-1+100-x\ge99\)
\(\left|x-2\right|+\left|99-x\right|\ge x-2+99-x\ge97\).............
\(\left|x-50\right|+\left|51-x\right|\ge x-50+51-x\ge1\)
\(\Rightarrow\left(\left|x-1\right|+\left|100-x\right|\right)+\left(\left|x-2\right|+\left|99-x\right|\right)+....+\left(\left|x-50\right|+\left|51-x\right|\right)\ge99+97+.....+3+1\)
\(\Rightarrow\left(\left|x-1\right|+\left|100-x\right|\right)+\left(\left|x-2\right|+\left|99-x\right|\right)+....+\left(\left|x-50\right|+\left|51-x\right|\right)\ge\dfrac{\left(99+1\right).50}{2}\)
\(\Rightarrow\left(\left|x-1\right|+\left|100-x\right|\right)+\left(\left|x-2\right|+\left|99-x\right|\right)+....+\left(\left|x-50\right|+\left|51-x\right|\right)\ge2500\)
Dấu "=" sảy ra khi:
\(\left\{{}\begin{matrix}x-50\ge0\\51-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge50\\x\le51\end{matrix}\right.\Rightarrow50\le x\le51\)
Vậy GTNN của biểu thức F là 2500 đạt được khi và chỉ khi \(50\le x\le51\)
Mình cũng không chắc đâu! Chúc bạn học tốt!!!