Думаю, что правильно, другого решения не вижу просто
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1) 6x^2-2x+5+10x-5x^2=x^2+3x+5
2) 6xy+8y-2xy-8y+1=4xy+1
<span>формула
![log_{a^k}b= \frac{1}{k}log_{a}b](https://tex.z-dn.net/?f=log_%7Ba%5Ek%7Db%3D+%5Cfrac%7B1%7D%7Bk%7Dlog_%7Ba%7Db+)
![log_{4}7=log_{2^2}7= \frac{1}{2}log_{2}7 \\ \\ log_{8}7=log_{2^3}7= \frac{1}{3}log_{2}7 \\ \\ log_{32}7=log_{2^5}7= \frac{1}{5}log_{2}7](https://tex.z-dn.net/?f=log_%7B4%7D7%3Dlog_%7B2%5E2%7D7%3D+%5Cfrac%7B1%7D%7B2%7Dlog_%7B2%7D7+%5C%5C++%5C%5C+log_%7B8%7D7%3Dlog_%7B2%5E3%7D7%3D+%5Cfrac%7B1%7D%7B3%7Dlog_%7B2%7D7+%5C%5C++%5C%5C+log_%7B32%7D7%3Dlog_%7B2%5E5%7D7%3D+%5Cfrac%7B1%7D%7B5%7Dlog_%7B2%7D7+)
Формула
![n\cdot log_{a}b=log_{a}b^n](https://tex.z-dn.net/?f=n%5Ccdot+log_%7Ba%7Db%3Dlog_%7Ba%7Db%5En+)
![log_{4}7=log_{2^2}7= \frac{1}{2}log_{2}7=log_{2}7^{ \frac{1}{2} }=log_ {2} \sqrt{7} \\ \\ log_{8}7=log_{2^3}7= \frac{1}{3}log_{2}7 =log_{2} \sqrt[3]{7} \\ \\ 2log_{32}7=2log_{2^5}7= 2\cdot \frac{1}{5}log_{2}7=log_{2}7^{ \frac{2}{5} }](https://tex.z-dn.net/?f=log_%7B4%7D7%3Dlog_%7B2%5E2%7D7%3D+%5Cfrac%7B1%7D%7B2%7Dlog_%7B2%7D7%3Dlog_%7B2%7D7%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D%3Dlog_+%7B2%7D+%5Csqrt%7B7%7D++%5C%5C++%5C%5C+log_%7B8%7D7%3Dlog_%7B2%5E3%7D7%3D+%5Cfrac%7B1%7D%7B3%7Dlog_%7B2%7D7+%3Dlog_%7B2%7D+%5Csqrt%5B3%5D%7B7%7D+%5C%5C+%5C%5C+2log_%7B32%7D7%3D2log_%7B2%5E5%7D7%3D+2%5Ccdot+%5Cfrac%7B1%7D%7B5%7Dlog_%7B2%7D7%3Dlog_%7B2%7D7%5E%7B+%5Cfrac%7B2%7D%7B5%7D+%7D)
Формулы
![log_{a}b-log_{a}c=log_{a} \frac{b}{c} \\ \\ log_{a} \frac{b}{c}+log_{a}d=log_{a} \frac{b}{c}\cdot d](https://tex.z-dn.net/?f=log_%7Ba%7Db-log_%7Ba%7Dc%3Dlog_%7Ba%7D+%5Cfrac%7Bb%7D%7Bc%7D+%5C%5C++%5C%5C+log_%7Ba%7D+%5Cfrac%7Bb%7D%7Bc%7D%2Blog_%7Ba%7Dd%3Dlog_%7Ba%7D+%5Cfrac%7Bb%7D%7Bc%7D%5Ccdot+d++)
log₄ 7-log₈ 7+2log₃₂ 7=(1/2)log₂7-(1/3)log₂7+2·(1/5)log₂7=
=log₂√7-log</span><span>₂∛7+log₂7²/⁵=log₂(√7/∛7)·7²</span><span>/⁵=log₂7¹⁷/³⁰</span>
Ответ:
2,3
Объяснение:
Решаем и подставляем:
3(5а-3b)-4(2a+7b)=15a-9b-8a-28b=7a-37b
При а=(-0,2), b=(-0,1)
7*(-0.2)-37*(-0,1)=-1,4+3,7=2,3