Zúp e câu này vs ạ
9x16 mũ X + 16x9 mũ X = 25x12 mũ x
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\(2^2\cdot3\left(x+5\right)-6^2=\left(2^3+2^2\right)\cdot2^2\)
\(\Rightarrow4\cdot\left(3x+15\right)-36=12\cdot4\)
\(\Rightarrow12x+60-36=48\)
\(\Rightarrow12x+24=48\)
\(\Rightarrow12x=24\)
\(\Rightarrow x=24:12=2\)
\(\left(x+6\right)^3=64\)
\(\Leftrightarrow\left(x+6\right)^3=4^3\)
\(\Leftrightarrow x+6=4\)
\(\Leftrightarrow x=-2\)
\(\left(x+6\right)^3=64\)
\(\sqrt[3]{\left(x+6\right)^3}=\sqrt[3]{64}\)
\(x+6=4\)
\(x=-2\)
`@` `\text {Ans}`
`\downarrow`
`(2^2+1) \times (x+14) = 5^2 \times 4 + (2^5 + 3^2 + 7^2) \div 2`
` \Rightarrow (4+1) \times (x+14) = 5^2\times 2^2 + ( 32 + 9 + 49) \div 2`
`\Rightarrow 5 \times (x+14) = (5*2)^2 + (32+58) \div 2`
`\Rightarrow 5 \times (x+14) = 10^2+90 \div 2`
`\Rightarrow 5 \times (x+14) = 100 + 45`
`\Rightarrow 5 \times (x+14) = 145`
`\Rightarrow x+14 = 145 \div 5`
`\Rightarrow x+14=29`
`\Rightarrow x=29-14`
`\Rightarrow x=15`
Vậy, `x=15.`
\(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
\(\Rightarrow\left(4+1\right)\cdot\left(x+14\right)=25\cdot4+\left(32+9+49\right):2\)
\(\Rightarrow5\cdot\left(x+14\right)=100+45\)
\(\Rightarrow5x+70=145\)
\(\Rightarrow5x=75\)
\(\Rightarrow x=\dfrac{75}{5}=15\)
a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)
\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)
\(=a^{12}\cdot b^3\)
b) \(b^6\cdot b\cdot c^7\cdot c^8\)
\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)
\(=b^7\cdot c^{15}\)
c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)
\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)
\(=a^{18}\cdot c^{21}\)
d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)
\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)
\(=a^5\cdot b^4\cdot c^4\)
a) Kiểm tra lại nhé
b) \(b^6.b^7.c^8\)
\(=b^{6+7}.c^8=b^{13}.c^8\)
c) \(a^8.a^9.a.c.c^{20}\)
\(=a^{8+9+1}.c^{1+20}\)
\(=a^{18}.c^{21}\)
d) \(a^2.a^3.b^4.c.c^3\)
\(=a^{2+3}.b^4.c^{1+3}\)
\(=a^5.b^4.c^4\)
\(#WendyDang\)
`4x^2 = 13`.
`=> x^2 = 13/4`.
`=> x = (sqrt 13)/(sqrt 4)`
`=> x = (+-sqrt 13)/2`.
Vậy `S = (+-sqrt 13)/2`.
\(2^{x+1}-2^x=3^2\)
\(\Rightarrow2^x\cdot\left(2-1\right)=9\)
\(\Rightarrow2^x=9\)
\(\Rightarrow x\in\varnothing\)
\(2^{x+1}-2^x=3^2\)
=>2^x*2-2^x=9
=>2^x=9
=>\(x\in\varnothing\)
b)\(3.4^2-2.3=3.\left(4^2-3\right)\)
\(=3.13\)
\(=39\)
Học tốt nha!!!
cô ơi cho con hỏi
11 mũ 5 chia cho 11 mũ n trừ 4 bằng 11 mũ 5
giúp tính bài giải cho con nhé
a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
\(9.16^x+16.9^x=25.12^x\)
\(\Leftrightarrow9.\left(\frac{16}{9}\right)^x+16=25.\left(\frac{12}{9}\right)^x\)
\(\Leftrightarrow9.\left(\frac{4}{3}\right)^{2x}-25.\left(\frac{4}{3}\right)^x+16=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\frac{4}{3}\right)^x=1=\left(\frac{4}{3}\right)^0\\\left(\frac{4}{3}\right)^x=\frac{16}{9}=\left(\frac{4}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)