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\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a/
\(\left(1-3x\right)^3=\left(-4\right)^3\Leftrightarrow1-3x=-4\Leftrightarrow x=\frac{5}{3}\)
b/
\(\left(4-3x\right)^4=\left(4-3x\right)^2\Leftrightarrow\left(4-3x\right)^2\left[\left(4-3x\right)^2-1\right]=0\)
\(\Leftrightarrow\left(4-3x\right)^2\left(5-3x\right)\left(3-3x\right)=0\)
\(\Leftrightarrow3-3x=0\) hoặc \(4-3x=0\) hoặc \(5-3x=0\)
\(\Leftrightarrow x=1\) hoặc \(x=\frac{4}{3}\) hoặc \(x=\frac{5}{3}\)
c/
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)\(\Leftrightarrow\frac{49.7^x+7.7^x+7^x}{57}=\frac{5^{2x}+5.5^{2x}+125.5^{2x}}{131}\)
\(\Leftrightarrow7^x=5^{2x}\Leftrightarrow7^x=25^x\Leftrightarrow\left(\frac{7}{25}\right)^x=1=\left(\frac{7}{25}\right)^0\)
\(\Rightarrow x=0\)
\(\left(1-3x\right)^3=-64\)
=> \(1-3x=-4\)
=> \(-3x=-4+1\) (chuyển vế)
=> \(-3x=-3\Rightarrow x=-3:\left(-3\right)=1\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
A = 2x2 - 3 . |x| + 1
+ Thay x = 2 vào biểu thức A, ta được:
A = 2 . 22 - 3 . |2| + 1
A = 2 . 4 - 3 . 2 + 1
A = 8 - 6 + 1
A = 2 + 1
A = 3
Vậy giá trị của biểu thức A tại x = 2 là 3.
Mình chỉ làm câu A thôi nhé bạn.
Chúc bạn học tốt!
a) C = 20013 - |5−2x|
do \(-\left|5-2x\right|\le0\forall x\)
=> 20013-\(\left|5-2x\right|\le20013\)
=>A≤20013
=> GTLN C =20013 khi 5-2x=0
=> 2x=5
=> x=\(\dfrac{5}{2}\)
vậy GTLN C = 20013 khi x=\(\dfrac{5}{2}\)
b) D = 7 - \(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\)
do \(-\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le0\forall x\)
=> 7-\(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le7\)
=> D≤7
=> GTLN D =7 khi \(\dfrac{2}{3}+\dfrac{1}{4}x=0\)
=> x=-\(\dfrac{8}{3}\)
Mình chỉ làm những câu rõ đề thôi nhé ^^
1/ a/ Đặt \(t=2x-3\) thì pt trở thành \(t^3=\left(t+2\right)^2\Leftrightarrow t^3-t^2-4t-4=0\Leftrightarrow t^2\left(t-1\right)-4\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^2-4\right)=0\Leftrightarrow\left(t-2\right)\left(t-1\right)\left(t+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}t=2\\t=1\\t=-2\end{array}\right.\)
Tới đây dễ rồi .
b/ Tương tự đặt \(a=2x-3\) thì pt trở thành \(a^3=a+2\Leftrightarrow a^3-a-2=0\)
Bạn xem lại đề , lớp 7 chưa học giải pt này đâu
c/ VT > 0 => VP > 0 => x > 0
Với x > 0 thì: \(\left|x+3\right|+\left|x+4\right|+\left|x+5\right|=x+3+x+4+x+5=3x+12\)
Tới đây dễ rồi :)
4) |2-|3-2x||=4
<=>\(\left[\begin{array}{nghiempt}2-\left|3-2x\right|=4\\2-\left|3-2x\right|=-4\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}\left|3-2x\right|=-2\left(vl\right)\\\left|3-2x\right|=6\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}3-2x=6\\3-2x=-6\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{9}{2}\end{array}\right.\)