Ai giải giúp mình 2 bài này với ạ
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Bài 3:
a: Hệ số tỉ lệ a là: \(a=7\cdot10=70\)
b: xy=70
=>\(y=\frac{70}{x}\)
c: Khi x=5 thì \(y=\frac{70}{5}=14\)
Khi x=14 thì \(y=\frac{70}{14}=5\)
d: Khi y=15 thì \(x=\frac{70}{y}=\frac{70}{15}=\frac{14}{3}\)
Khi y=3 thì \(x=\frac{70}{y}=\frac{70}{3}\)
Bài 4:
a: Xét ΔMNI và ΔMPI có
MN=MP
NI=PI
MI chung
Do đó: ΔMNI=ΔMPI
b: ΔMNI=ΔMPI
=>\(\hat{MNI}=\hat{MPI}\)
c: ΔMNI=ΔMPI
=>\(\hat{NMI}=\hat{PMI}\)
=>MI là phân giác của góc NMP
d: ΔMNI=ΔMPI
=>\(\hat{MIN}=\hat{MIP}\)
mà \(\hat{MIN}+\hat{MIP}=180^0\) (hai góc kề bù)
nên \(\hat{MIN}=\hat{MIP}=\frac{180^0}{2}=90^0\)
=>MI⊥NP
mà I là trung điểm của NP
nên MI là đường trung trực của NP
Bài 3:
a. \(R=R1+R2=15+30=45\Omega\)
b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)
Bài 4:
\(I1=U1:R1=6:3=2A\)
\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)
\(U=R.I=\left(3+15\right).2=36V\)
\(U2=R2.I2=15.2=30V\)
`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`
`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`
`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`
`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`
`<=> cos2x - cos4x - cos2x = 1/2`
`<=> cos4x = cos(2π)/3`
`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)
\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)
Ta lấy vễ trên chia vế dưới
\(=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)
Ta lấy vế trên chia vế dưới
\(=2^3.3=24\)
\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)







Bài 1:
a: \(-\frac52-\frac49\cdot\frac{21}{16}\)
\(=-\frac52-\frac{4}{16}\cdot\frac{21}{9}\)
\(=-\frac52-\frac14\cdot\frac73=-\frac52-\frac{7}{12}=\frac{-30-7}{12}=-\frac{37}{12}\)
b: \(\frac{-5}{4}\cdot\frac{-3}{7}+\frac{-5}{4}\cdot\frac{17}{7}+12\frac12\)
\(=-\frac54\left(-\frac37+\frac{17}{7}\right)+12,5\)
\(=-\frac54\cdot\frac{14}{7}+12,5=-\frac54\cdot2+12,5=-\frac52+12,5=10\)
c: \(\left(-\frac13\right)^2+60\%+2\frac89-\frac85\)
\(=\frac19+\frac35+2+\frac89-\frac85\)
\(=2+\left(\frac19+\frac89\right)+\left(\frac35-\frac85\right)=2\)
d: \(15\frac37-\left(\frac89+7\frac37\right)\)
\(=15+\frac37-\frac89-7-\frac37\)
\(=8-\frac89=\frac{72-8}{9}=\frac{64}{9}\)
Bài 2:
a: \(1,75-20\%\cdot x=\frac15\)
=>\(0,2x=1,75-0,2=1,55\)
=>\(x=1,55:0,2=1,55\cdot5=7,75\)
b: \(\left(\frac23-x\right):\frac{-3}{10}=75\%\)
=>\(\frac23-x=\frac34\cdot\frac{-3}{10}=\frac{-9}{40}\)
=>\(x=\frac23+\frac{9}{40}=\frac{80+27}{120}=\frac{107}{120}\)
c: \(\left|\frac34x-\frac15\right|=0,25\)
=>\(\left[\begin{array}{l}\frac34x-\frac15=\frac14\\ \frac34x-\frac15=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac34x=\frac15+\frac14=\frac{9}{20}\\ \frac34x=-\frac14+\frac15=\frac{-1}{20}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{9}{20}:\frac34=\frac{9}{20}\cdot\frac43=\frac{36}{60}=\frac35\\ x=-\frac{1}{20}:\frac34=-\frac{1}{20}\cdot\frac43=\frac{-4}{60}=-\frac{1}{15}\end{array}\right.\)
d: \(x-\frac29x=\frac59\)
=>\(x\left(1-\frac29\right)=\frac59\)
=>\(x\cdot\frac79=\frac59\)
=>\(x=\frac59:\frac79=\frac57\)