Ответы
Ответ дал:
0
1
1)x<0
(x²+14)/3=-3x
x²+14+9x=0
x1+x2=-9 U x1*x2=14
x1=-7 U x2=-2
2)x≥0
x²-9x+14=0
x1+x2=9 U x1*x2=14
x1=2 U x2=7
Ответ x={-7;-2;2;7}
2
1)√(ab)-ab/[√a(√a+√b)]=√(ab)-b√a/(√a+√b)=(a√b+b√a-b√a)/(√a+√b)=
=a√b/(√a+√b)
2)a√b/(√a+√b)*(√a-√b)(√a+√b)/![[ sqrt[4]{b} ( sqrt[4]{a} - sqrt[4]{b} )=a sqrt[4]{b} ( sqrt[4]{a} - sqrt[4]{b} )( sqrt[4]{a} + sqrt[4]{b})/( sqrt[4]{a} - sqrt[4]{b} )= [ sqrt[4]{b} ( sqrt[4]{a} - sqrt[4]{b} )=a sqrt[4]{b} ( sqrt[4]{a} - sqrt[4]{b} )( sqrt[4]{a} + sqrt[4]{b})/( sqrt[4]{a} - sqrt[4]{b} )=](https://tex.z-dn.net/?f=%5B+sqrt%5B4%5D%7Bb%7D+%28+sqrt%5B4%5D%7Ba%7D+-+sqrt%5B4%5D%7Bb%7D+%29%3Da+sqrt%5B4%5D%7Bb%7D+%28+sqrt%5B4%5D%7Ba%7D+-+sqrt%5B4%5D%7Bb%7D+%29%28+sqrt%5B4%5D%7Ba%7D+%2B+sqrt%5B4%5D%7Bb%7D%29%2F%28+sqrt%5B4%5D%7Ba%7D+-+sqrt%5B4%5D%7Bb%7D+%29%3D)
![a sqrt[4]{b} ( sqrt[4]{a} + sqrt[4]{b} ) a sqrt[4]{b} ( sqrt[4]{a} + sqrt[4]{b} )](https://tex.z-dn.net/?f=a+sqrt%5B4%5D%7Bb%7D+%28+sqrt%5B4%5D%7Ba%7D+%2B+sqrt%5B4%5D%7Bb%7D+%29)
3
4*2^x-4/2^x-15=0
2^x=a
4a²-15a-4=0
D=225+64=289
a1=(15-17)/8=-1/4⇒2%x=-1/4 нет решения
a2=(15+17)/8=4⇒2&x=4⇒x=2
4
sina-cosa=n
sinacosa=1/2*[sin²a+cos²a-(sina-cosa)²]=1/2(1-n²)
sin³a-cos³a=(sina-cosa)(sin²a+sinacosa+cos²a)=
=n*(1+1/2*(1-n²))=n(3/2-n²/2)
6
ОДЗ
{x-2>0⇒x>2
{x-4≠0⇒x≠4
x∈(2;4) U (4;∞)
log(3)[(x-2)(x-4)]²=0
[(x-2)(x-4)]²=1
x²-6x+8=-1 U x²-6x+8=1
x²-6x+9=0⇒(x-3)²=0⇒x=3
x²-6x+7=0
D=36-28=8
x1=(6-2√7)/2=3-√7 не удов усл
x2=3+√7
8
(x+2)(x-3)/(x+6)>0
x=-2 x=3 x=-6
_ + _ +
-----------(-6)----------(-2)----------------(3)-----------------------------
x∈(-6;-2) U (3;∞)
1)x<0
(x²+14)/3=-3x
x²+14+9x=0
x1+x2=-9 U x1*x2=14
x1=-7 U x2=-2
2)x≥0
x²-9x+14=0
x1+x2=9 U x1*x2=14
x1=2 U x2=7
Ответ x={-7;-2;2;7}
2
1)√(ab)-ab/[√a(√a+√b)]=√(ab)-b√a/(√a+√b)=(a√b+b√a-b√a)/(√a+√b)=
=a√b/(√a+√b)
2)a√b/(√a+√b)*(√a-√b)(√a+√b)/
3
4*2^x-4/2^x-15=0
2^x=a
4a²-15a-4=0
D=225+64=289
a1=(15-17)/8=-1/4⇒2%x=-1/4 нет решения
a2=(15+17)/8=4⇒2&x=4⇒x=2
4
sina-cosa=n
sinacosa=1/2*[sin²a+cos²a-(sina-cosa)²]=1/2(1-n²)
sin³a-cos³a=(sina-cosa)(sin²a+sinacosa+cos²a)=
=n*(1+1/2*(1-n²))=n(3/2-n²/2)
6
ОДЗ
{x-2>0⇒x>2
{x-4≠0⇒x≠4
x∈(2;4) U (4;∞)
log(3)[(x-2)(x-4)]²=0
[(x-2)(x-4)]²=1
x²-6x+8=-1 U x²-6x+8=1
x²-6x+9=0⇒(x-3)²=0⇒x=3
x²-6x+7=0
D=36-28=8
x1=(6-2√7)/2=3-√7 не удов усл
x2=3+√7
8
(x+2)(x-3)/(x+6)>0
x=-2 x=3 x=-6
_ + _ +
-----------(-6)----------(-2)----------------(3)-----------------------------
x∈(-6;-2) U (3;∞)
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