giup minh voi cac ban

\(2\cdot2^2\cdot2^3\cdot2^4\cdot\cdot\cdot2^x=32768\)

\(\Leftrightarrow2^{1+2+3+4+\cdot\cdot\cdot+x}=2^{15}\)

\(\Leftrightarrow1+2+3+4+..+x=15\)

\(\Leftrightarrow\)\(\frac{\left(1+x\right)x}{2}=15\)

\(\Leftrightarrow x\left(x+1\right)=30=5\left(5+1\right)\)

Vậy x=5

Bài 2:

Bậc của đơn thức là 2+5+3=10

Bài 3:

\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=5\)

+)TH1: \(x\ge\frac{1}{4}\) thì bt trở thành

\(2x-\frac{1}{2}=5\Leftrightarrow2x=\frac{11}{2}\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)

+)TH2: \(x< \frac{1}{4}\) thì pt trở thành

\(2x-\frac{1}{2}=-5\Leftrightarrow2x=-\frac{9}{2}\Leftrightarrow x=-\frac{9}{4}\left(tm\right)\)

Vậy x={-9/4;11/4}

2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)

\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3

3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)

4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)

1. Đặt \(t=x^2,t\ge0\)

\(3x^4+4x^2-2\ge3.0+4.0-2=-2\)

=> MIN = -2 khi x = 0

2. \(\left(x^2+2\right)\left(x+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+2=0\\x+1=0\end{array}\right.\)

Vì \(x^2+2\ge2>0\) => Vô nghiệm

Vậy x+1 = 0 => x = -1

3. Kết quả là 10

4. Ko rõ đề

\(\left(x+y\right)=3\Leftrightarrow\left(x+y\right)^2=9\Leftrightarrow x^2+y^2+2xy=9\Leftrightarrow5+2xy=9\Leftrightarrow xy=2.\)

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=3.\left(5-2\right)=9\)

Câu 6:

\(\left(x-2016\right)^2\ge0\) với mọi x

\(\left(x+2017\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-2016\right)^2+\left(y+2017\right)^2=0\) Khi \(\left(x-2016\right)^2=0\Leftrightarrow x=2016\) và \(\left(x+2017\right)^2=0\Leftrightarrow x=-2017\)

\(\Rightarrow x+y=2016-2017=-1\)

Câu 7:

 \(D=\left(x+y\right)^2-6\left(x+y\right)-15=\left(-9\right)^2-6.\left(-9\right)-15=120\)

\(Q=\left(x+y\right)^2-4\left(x+y\right)+1=3^2-4.3+1=-2\)

câu 5:

x²+y²=5   -> x²+2xy+ y²-2xy=5

                -> (x+y)² - 2xy = 5 -> 3²  - 2xy = 5 ->xy = 2

có x³+ y³= (x+y).(x²-xy+y²)

              = 3.( 5- 2)= 9

vậy x³+ y³ =9

câu 6:

( x - 2016)²  ≥ 0 dấu = xảy ra khi x=2016

 ( y + 2017 )²  ≥ 0 dấu bằng xảy ra khi y = 2016

-> ( x - 2016)² + ( y + 2017 )²  ≥ 0 dấu bằng xảy ra khi x=2016, y = 2017

-> x+y=2016+2017=4033

câu 7:

a,

D = x² +2xy +y²  - 6x - 6y  -15= (x² +2xy +y²⁾  - (6x + 6y)  -15= (x+y)² - 6(x+y) - 15

D= (-9)² -6.(-9)-15=120

b,

Q = x² + 2xy + y²  - 4x - 4y +1 = (x² + 2xy + y²⁾  - (4x + 4y) +1

Q= (x+y)²-4.(x+y)+1

Q=3²- 4.3 +1= -2

5.\(C\text{ó}x^2-12=0\Rightarrow x^2=12\Rightarrow x=\sqrt{12}ho\text{ặc}x=-\sqrt{12}\)

Mà x>0\(\Rightarrow x=\sqrt{12}\)

6.Vì x-y=4\(\Rightarrow\left(x-y\right)^2=x^2-2xy+y^2=x^2-10+y^2=4^2=16\Rightarrow x^2+y^2=26\)

Có \(\left(x+y\right)^2=x^2+2xy+y^2=26+10=36=6^2=\left(-6\right)^2\)

Vì xy>0 và x>0 =>y>0=>x+y>0=>x+y=6

7. \(3x^2+7=\left(x+2\right)\left(3x+1\right)\)

\(3x^2+7=3x^2+7x+2\)

\(3x^2+7-3x^2-7x-2=0\)

-7x+5=0

-7x=-5

\(x=\frac{5}{7}\)

8.\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left(2x+4\right)^2=9\)

(2x+1-2x-4)(2x+1+2x+4)=9

-3(4x+5)=9

4x+5=-3

4x=-8

x=-2

Còn câu 9 và 10 để mình nghiên cứu đã

biet x+y =2 tinh min 3x^2 + y^2

x² +4x+y²-12 =0 => (x+2)² =(4-y)(4+y) ; vì x;y thuộc Z => 4-y = 4+y => y =0 => (x+2)² =16

x +2 = 4 => x =2 

hoăc x+2 =-4 => x =-6

=> Pmax=x² +y² = (-6)² +0 = 36 khi x = -6; y =0