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1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
1;Ta có\(5.3^x=5.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
2.Ta có \(9.5^x=6.5^6+3.5^6\)
\(\Rightarrow9.5^x=5^6.\left(6+3\right)\)
\(\Rightarrow9.5^x=9.5^6\)
\(\Rightarrow5^x=5^6\)\
\(\Rightarrow x=6\)
3, Ta có \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
\(\Rightarrow3^{x+1}.\left(2.3+4\right)=10.3^6\)
\(\Rightarrow3^{x+1}.10=10.3^6\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
a) 5.3x = 5.34
=> 3x=34
=> x=4
b) 9.5x=6.56+3.56
=> 9.5x = (6+3)56
=> 9.5x=9.56
=> 5x=56
=> x=6
c) 2.3x+2 + 4.3x+1 = 10.36
=> 2.3x+1.3 + 4.3x+1 = 10.36
=> 6.3x+1+4.3x+1=10.36
=> (6+4).3x+1=10.36
=> 10.3x+1=10.36
=> 3x+1=36
=> x+1=6
=> x=5
a,?????
b, Với mọi giá trị của x;y ta có:
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\)
Để \(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..........
c, \(\left|2x\right|-\left|3,5\right|=\left|-6,5\right|\)
\(\Rightarrow\left|2x\right|=6,5+3,5=10\)
\(\Rightarrow\left\{{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy..........
d, \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\left\{{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{z}{\dfrac{1}{2}}=\dfrac{x-z}{\dfrac{2}{3}-\dfrac{1}{2}}=\dfrac{-13}{2}:\dfrac{1}{6}=-39\)
Do đó: x=-26; y=-117/5; z=-39/2
Câu hỏi của Quách Quỳnh Bảo Ngọc - Toán lớp 7 - Học toán với OnlineMath.Em tham khảo cách làm ở link này nhé!
đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)
=> x.y=5k.7k=35k2=875
k2=875:35=25
<=>k2=52
k={-5,5}
Thay k :
\(\hept{\begin{cases}x=5k=5.5=25\\y=7k=7.5=35\end{cases}}\) hoặc \(\hept{\begin{cases}x=5k=5.\left(-5\right)=-25\\y=7k=7.\left(-5\right)=-35\end{cases}}\)
Vậy \(x=\pm\)25;y=\(\pm\)35
a: \(\dfrac{5^5}{5^x}=5^{18}\)
=>5-x=18
hay x=-13
b: \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)
=>4-x=50
hay x=-46
c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)
=>2x-3=49
=>2x=52
hay x=26
d: \(\dfrac{2^3}{2^x}=4^5\)
\(\Leftrightarrow2^{3-x}=2^{10}\)
=>3-x=10
hay x=-7
e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
f: \(7\cdot2^x=2^9+5\cdot2^8\)
\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)
\(\Leftrightarrow2^x=2^8\)
hay x=8
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
\(5^{x+1}+6.5^{x+1}=875\)
\(\Rightarrow5^{x+1}.\left(1+6\right)=875\)
\(\Rightarrow5^{x+1}.7=875\)
\(\Rightarrow5^{x+1}=875:7=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1=2\)