1: \(-1\le\sin2x\le1\)

=>\(2\ge-2\cdot\sin2x\ge-2\)

=>2+3>=-2sin2x+3>=-2+3

=>5>=y>=1

y max=5 khi sin 2x=-1

=>\(2x=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac{\pi}{4}+k\pi\)

y min=1 khi sin 2x=1

=>\(2x=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{4}+k\pi\)

2: \(-1\le cos7x\le1\)

=>-4<4cos7x<=4

=>-4-2<=4*cos7x-2<=4-2

=>-6<=y<=2

y min=-6 khi cos7x=-1

=>\(7x=\pi+k2\pi\)

=>\(x=\frac{\pi+k2\pi}{7}\)

y max=2 khi cos7x=1

=>\(7x=k2\pi\)

=>\(x=\frac{k2\pi}{7}\)

3: \(0\le\sin^24x\le1\)

=>\(0\ge-\sin^24x\ge-1\)

=>\(0+1\ge-\sin^24x+1\ge-1+1\)

=>\(1\ge-\sin^24x+1\ge0\)

=>\(\frac15\ge\frac{-\sin^24x+1}{5}\ge0\)

y max=1/5 khi \(\sin^24x=0\)

=>sin 4x=0

=>\(4x=k\pi\)

=>\(x=\frac{k\pi}{4}\)

y min=0 khi \(\sin^24x=1\)

=>\(cos^24x=0\)

=>cos4x=0

=>\(4x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{8}+\frac{k\pi}{4}\)

6: \(y=\sin x+cosx+2\)

\(=\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)+2\)

Ta có: \(-1\le\sin\left(x+\frac{\pi}{4}\right)\le1\)

=>\(-\sqrt2\le\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)\le\sqrt2\)

=>\(-\sqrt2+2\le\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)+2\le\sqrt2+2\)

\(y_{\min}=-\sqrt2+2\) khi \(\sin\left(x+\frac{\pi}{4}\right)=-1\)

=>\(x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)

=>\(x=-\frac34\pi+k2\pi\)

y max=\(\sqrt2+2\) khi \(\sin\left(x+\frac{\pi}{4}\right)=1\)

=>\(x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{4}+k2\pi\)

\(7x+\left(-6\right)=0\\ \Leftrightarrow7x=6\\ \Leftrightarrow x=\dfrac{6}{7}\)

Vậy nghiệm của đa thức p(x) là \(x=\dfrac{6}{7}\)

Đa thức \(P\left(x\right)\) có nghiệm khi: 

\(P\left(x\right)=0\)

\(\Rightarrow7x+\left(-6\right)=0\)

\(\Rightarrow7x-6=0\)

\(\Rightarrow7x=6\)

\(\Rightarrow x=\dfrac{6}{7}\)

Vậy nghiệm của đa thức \(P\left(x\right)\) là \(\dfrac{6}{7}\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

=>5x^2-6x-11=0

=>5x^2-11x+5x-11=0

=>(5x-11)(x+1)=0

=>x=11/5 hoặc x=-1

Bài 1 :

a, \(A=x\left(x-6\right)+10\)

=x^2 - 6x + 10

=x^2 - 2.3x+9+1

=(x-3)^2 +1 >0 Với mọi x dương

Cảm ơn bạn Vũ Anh Quân ;) ;) ;) 

\(B=\left(x-8x-3\right)\)

\(B=\left(x^2-2x4-16\right)+13\)

\(-B=\left(x^2+2x4+16\right)-13\)

\(-B=\left(x+4\right)^2-13\ge-13\)

\(B=-\left(x+4\right)^2+13\le13\)

Dấu "=" xảy ra khi và chỉ khi \(-\left(x+4\right)^2=0\)

\(\Leftrightarrow\left(x+4^2\right)=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy GTLN của B là 13 khi và chỉ khi x=-4

Sai môn học r bạn ơi

a)x\(\in\left\{-5;-4;-3;-2;-1\right\}\)

b)x\(\in\left\{-2;-1;0;1;2;3\right\}\)

c)x\(\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

d)x\(\in\left\{1;2;3;4\right\}\)

e)x\(\in\left\{0;1;2;3;4\right\}\)

f)x\(\in\left\{-10;-9;-8;-7\right\}\)

Ta có: \(\left(x^{3n}+y^{3n}\right)\left(x^{3n}-y^{3n}\right)=-x^{6n}-y^{6n}\)

\(\Leftrightarrow x^{6n}-y^{6n}=-x^{6n}-y^{6n}\)

\(\Leftrightarrow n\in\varnothing\)

\(\dfrac{x-2}{5}=\dfrac{1-x}{6}\\ =>\left(x-2\right)\cdot6=\left(1-x\right)\cdot5\\ =>6x-12=5-5x\\ =>6x+5x=5+12\\ =>11x=17\\ x=\dfrac{17}{11}\)

`[x-2]/5=[1-x]/6`

`=>6(x-2)=5(1-x)`

`=>6x-12=5-5x`

`=>6x+5x=5+12`

`=>11x=17`

`=>x=17/11`