tìm X biết
\(5^x.\left(5^3\right)^2=625\)
\(27< 81^3:3^x< 243\)
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
giúp mình vs nha
bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
1.Viết dưới dạng thu gọn
a, \(8^5.64^2.16^3\)
b, \(81^7.243^3.3^{11}.9^7\)
c. \(6^{15}.8^{17}.9^{31}.81^31024^5\)
2. tìm x biết
a, \(\frac{4}{15}:x+\frac{-1}{3}=\frac{5}{9}\)
b,\(\left(3,25-\frac{2}{3}x\right):\frac{-7}{4}=-3\)
c, \(71+\left(26-3x\right):5=75\)
d, \(3+2^{x-1}=24-[4^2-\left(2^2-1\right)]\)
e, \(\frac{5}{7}-\left|3x+1\right|=\frac{-2}{7}\)
3.tính
a, \(\left(\frac{2}{3}\right)^2:\left(-8\right)-\left(\frac{-1}{4}+\frac{9}{20}\right):1\frac{4}{5}\)
b,\(0,8.\frac{-15}{14}-\frac{4}{5}-\frac{13}{14}-1\frac{2}{5}\)
CÁC BẠN GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
THANKS
Tĩm x:
1,\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+5}\)
2,\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}+......+x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Giải chi tiết hộ mình nha mình tik cho
Bài 1 : Tính
\(a,\left[\frac{-54}{64}-\frac{1}{9}.\frac{8}{27}.\frac{-1}{3}\right].\frac{-81}{128}\)
\(b,\left[\left(0,1\right)^2\right]^0+7^2.\frac{1}{49}.\left[\left(2^2\right)^3:2^5\right]\)
Các bạn giúp mình với mình đang cần gấp
bài1: tìm x:
a)\(8< 2^x< =2^9.2^5\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\frac{2}{5}\right)^x\left(\frac{5}{2}\right)^{-3}.\left(\frac{-2}{5}\right)^2\)
d)\(\left(5x+1\right)^2=\frac{36}{49}\)
e)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\) f)\(\left(8x-1\right)^{2n+1}=5^{2n+1}\)(n thuộc N)
bài 2:tìm x,y biết:
a)\(x^2+\left(y-\frac{1}{10}\right)^4=0\)
b)\(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}< =6\)
c)\(\left(x-7\right)^{x+1}-\left(x-y\right)^{x+11}=0\)
bài 3:tìm giá trị nhỏ nhất:
\(A=\left(2x+\frac{1}{3}\right)^2-1\)
tìm Gía trị lớn nhất :\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)
baif4: tìm x,y:
\(x.\left(x-y\right)=\frac{1}{10}\) \(\)
giúp mình với nhé
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
\(a,\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(b,\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(c,\left(-3\right)^{x+5}=\frac{1}{81}\)
\(d,\left(\frac{1}{9}^x\right)=\left(\frac{1}{27}\right)^6\)
\(e,\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(f,5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(r,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(h,\left(\frac{1}{2}-\frac{1}{3}\right).6x+6^{x+2}=6^{10}+6^7\)
nhờ mấy bn giúp mk tối mình nộp rồi
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)
\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)
\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Bài 1:Tìm x
a)\(\left(5x+1\right)^2=\frac{36}{49}\)
b)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
c)\(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
a) (5x + 1)2 = 36/49
<=> (5x + 1)2 = (6/7)2
<=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
<=>\(\orbr{\begin{cases}5x=\frac{6}{7}-1\\5x=-\frac{6}{7}-1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
Vậy: ...
tìm x biết
a, ( 2x - 3 ) ( x + 1 ) <0
b, ( x - \(\frac{1}{2}\) ) ( x + 3) >0
c,\(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
biết không thuộc { -2, -5 ,-10 ,-17 }
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)