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\(a,\left|5x+4\right|+7=26\\ \left|5x+4\right|=26+7\\ \left|5x+4\right|=33\\ \Rightarrow\left\{{}\begin{matrix}5x+4=33\\5x+4=-33\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x=29\\5x=-29\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\\x=-\dfrac{29}{5}\end{matrix}\right.\)
Các câu sau làm tương tự!
a) \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)
<=>\(\hept{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{4}\end{cases}}}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}}\)
\(\left(2x-1\right)\left(3x-2\right)-\left(6x-1\right)\left(x-3\right)=17\)
\(\Rightarrow\left(6x^2-4x-3x+2\right)-\left(6x^2-18x-x+3\right)=17\)
\(\Rightarrow6x^2-4x-3x+2-6x^2+18x+x-3=17\)
\(\Rightarrow\left(6x^2-6x^2\right)+\left(18x-4x-3x+x\right)-\left(3+2\right)=17\)
\(\Rightarrow12x-1=17\)
\(\Rightarrow12x=18\)
\(\Rightarrow x=\frac{3}{2}\)
6x2 - 4x - 3x + 2 - 6x2 - 18x - x + 3 = 17
=>>> -27x + 5 = 17 =>>> -27x = 12 =>>> x = -4/7
Vậy: x = -4/7
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
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 \(x=\frac{1}{2};y=3\)
a.\(\left(\left(\frac{3}{4}\right)^3\right)^2=\left(\frac{16}{9}\right)^x\Leftrightarrow\left(\frac{3}{4}\right)^6=\left(\frac{4}{3}\right)^{2x}\Leftrightarrow x=-3\)
b. \(\left(\frac{1}{3}\right)^x=3^{-3}\Leftrightarrow\left(\frac{1}{3}\right)^x=\left(\frac{1}{3}\right)^3\Leftrightarrow x=3\)
a) I 5x+4I +7=26 b) 3 I 9-2xI - 17=16
I 5x+4 I = 26-7 3 I 9-2xI=16+17
I 5x+4 I =19 3 I 9-2xI=33
=> 5x+4=19 hoặc 5x+4=-19 I 9-2xI=33:3=11
5x = 19-4=15 hoặc 5x=-19-4=-23 => 9-2x=11 hoặc 9-2x=-11
-2x=11-9=2 hoặc -2x=-11+9=-2
x=2:(-2)=-1 hoặc x=-2:(-2)=1
a) \(\left|5x+4\right|+7=26\)
\(\Rightarrow\left|5x+4\right|=26-7\)
\(\Rightarrow\left|5x+4\right|=19\)
\(\Rightarrow\orbr{\begin{cases}5x+4=19\\5x+4=-19\end{cases}\Rightarrow\orbr{\begin{cases}5x=19-4\\5x=-19-4\end{cases}\Rightarrow}\orbr{\begin{cases}5x=15\\5x=-23\end{cases}\Rightarrow}\orbr{\begin{cases}x=15:5\\x=-23:5\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-4,6\end{cases}}}\)
Vậy \(x\in\left\{3;-4,6\right\}\)
b) \(3\left|9-2x\right|-17=16\)
\(\Rightarrow3\left|9-2x\right|=16+17\)
\(\Rightarrow3\left|9-2x\right|=23\)
\(\Rightarrow\left|9-2x\right|=23:3\)
\(\Rightarrow\left|9-2x\right|=\frac{23}{3}\)
\(\Rightarrow\orbr{\begin{cases}9-2x=\frac{23}{3}\\9-2x=-\frac{23}{3}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{23}{3}+9\\2x=-\frac{23}{3}+9\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{23}{3}+\frac{27}{3}\\2x=-\frac{23}{3}+\frac{27}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{50}{3}\\2x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{50}{3}:3\\x=4:2\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{50}{3}\times\frac{1}{3}\\x=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{50}{9}\\x=2\end{cases}}}\)
Vậy \(x\in\left\{\frac{50}{9};4\right\}\)
Chúc bạn học tốt!