x:1/4=8
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\(\left(\dfrac{2x}{3}-\dfrac{1}{3}\right)+\left(3x-2x+1\right)=8\)
\(\Leftrightarrow\dfrac{2x-1}{3}+x-7=0\Rightarrow2x-1+3x-21=0\Leftrightarrow x=\dfrac{22}{5}\)
\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left[3x-2\left(x-1\right)\right]=8\)
\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+2=8\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{5}\)
a: \(\Leftrightarrow3x+9=-2x+6\)
=>5x=-3
hay x=-3/5
b: =>3/x=y/35=3/7
=>x=7; y=15
c: =>9x/5=-3/5
=>9x=-3
hay x=-1/3
d: =>x+2/26=-1/4
=>x+2=-13/2
hay x=-17/2
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\) và \(x^2-2y^2+z^2=8\)
Áp dụng t/c dãy tsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2y^2}{18}=\dfrac{z^2}{16}=\dfrac{x^2-2y^2+z^2}{4-18+16}=\dfrac{8}{2}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\\z^2=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\\z=\pm8\end{matrix}\right.\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
Ta có: \(x^2-2y^2+z^2=8\)
\(\Leftrightarrow4k^2-18k^2+16k^2=8\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=4k=8\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=4k=-8\end{matrix}\right.\)
ĐK : \(x\ne-2.-3;-4;-5;-6\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )
Vậy tập nghiệm phương trình là S = { -10 ; 2 }
ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)
Phương trình tương đương với:
\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)
b) Gọi giao điểm của (d) với Ox là điểm A. \(\Rightarrow y=0.\)
\(\Rightarrow\) \(OA=\left|\dfrac{-4}{m}\right|=\dfrac{4}{\left|m\right|}.\) (đvđd).
Gọi giao điểm của (d) với Oy là điểm B. \(\Rightarrow x=0.\)
\(\Rightarrow OB=4\) (đvđd).
Ta có: \(S_{\Delta ABC}=\dfrac{1}{2}OA.OB=\dfrac{1}{2}.\dfrac{4}{\left|m\right|}.4=8\) (đvdt).
\(\Rightarrow\dfrac{4}{\left|m\right|}=4.\Leftrightarrow\left|m\right|=1.\Leftrightarrow\left[{}\begin{matrix}m=1.\\m=-1.\end{matrix}\right.\)
a, Điều kiện: x > 0
\(log_3\left(x\right)< 2\\ \Rightarrow0< x< 9\)
b, Điều kiện: x > 5
\(log_{\dfrac{1}{4}}\left(x-5\right)\ge-2\\ \Rightarrow x-5\le16\\ \Leftrightarrow5< x\le21\)
Dấu ngoặc và cuối là sai nhé bạn. Phải là ngoặc vuông (x=0 hoặc x=-8) mới đúng, vì x không thể nhận 2 giá trị khác nhau cùng lúc.
=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2
Đặt x+1/x=a(a>=2)
=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2
=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2
=>(x+4)^2=16
=>x+4=4 hoặc x+4=-4
=>x=-8;x=0
`@` `\text {Ans}`
`\downarrow`
`9^8 \div 3^2`
`= (3^2)^8 \div 3^2`
`= 3^16 \div 3^2`
`=`\(3^{16-2}=3^{14}\)
_____
`3^7 * 27^5 * 81^3`
`= 3^7*(3^3)^5 * (3^4)^3`
`= 3^7 * 3^15 * 3^12`
`=`\(3^{7+15+12}\)
`= 3^34`
______
`36^5 \div 18^5`
`= (36 \div 18)^5`
`= 2^5 = 32`
______
`24*5^5 + 5^2*5^3`
`= 24*5^5 + 5^5`
`= 5^5*(24+1)`
`= 5^5 * 25`
`= 5^5*5^2`
`= 5^7`
______
`125^4 \div 5^8`
`= (5^3)^4 \div 5^8`
`= 5^12 \div 5^8`
`= 5^4`
_____
`@` Phép nâng lên lũy thừa: \(\left(a^m\right)^n=a^{m\cdot n}\)
`@` Chia lũy thừa cùng cơ số: \(a^m\div a^n=a^{m-n}\)
`@` Nhân lũy thừa cùng cơ số: \(a^m\cdot a^n=a^{m+n}\)
\(9^8:3^2\)
\(=\left(3^2\right)^8:3^2\)
\(=3^{16}:3^2\)
\(=3^{14}\)
=============
\(3^7\cdot27^5\cdot81^3\)
\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)
\(=3^7\cdot3^{15}\cdot3^{12}\)
\(=3^{7+12+15}\)
\(=3^{34}\)
===============
\(36^5:18^5\)
\(=\left(36:18\right)^5\)
\(=2^5\)
\(=32\)
=============
\(24\cdot5^5+5^2\cdot5^3\)
\(=24\cdot5^5+5^5\)
\(=5^5\cdot\left(24+1\right)\)
\(=5^5\cdot25\)
\(=5^5\cdot5^2\)
\(=5^7\)
==============
\(125^4:5^8\)
\(=\left(5^3\right)^4:5^8\)
\(=5^{12}:5^8\)
\(=5^4\)
\(4^{x+3}+4^{x+2}+4^{x+1}+4^x=5440\)
\(\Rightarrow4^x.4^3+4^x.4^2+4^x.4+4^x=5440\)
\(\Rightarrow4^x\left(4^3+4^2+4+1\right)=5440\)
\(\Rightarrow4^x.\left(64+16+4+1\right)=5440\)
\(\Rightarrow4^x.85=5440\)
\(\Rightarrow4^x=5440:85\)
\(\Rightarrow4^x=64=4^3\)
\(\Rightarrow x=3\)
dễ quá bạn ơi giải câu này nè mới chất
Q= 12 + 22 + 32 +...+ 1002
\(x\div\frac{1}{4}=8\)
\(x=8\times\frac{1}{4}\)
\(x=\frac{8}{4}\)
\(x=2\)
x;1/4=8
x=8;1/4
x=2