Ôn tập toán 8 - Hỏi đáp

Bài 2:

a) \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\Rightarrow\left(\dfrac{x-17}{33}-1\right)+\left(\dfrac{x-21}{29}-1\right)+\left(\dfrac{x}{25}-2\right)=0\)

\(\Rightarrow\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\right)=0\)

Mà \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\ne0\)

\(\Rightarrow x-50=0\)

\(\Rightarrow x=50\)

Vậy x = 50

Bải 3a

\(\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{-a}{2a}+\dfrac{b+c}{2a}+\dfrac{-b}{2b}+\dfrac{c+a}{2b}+\dfrac{-c}{2c}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)

\(\Leftrightarrow-\dfrac{3}{2}+\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge3\)

\(\Leftrightarrow\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge6\)

\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge6\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2+2+2=6\)

\(\Leftrightarrow\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

Bài 3b)

\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)

\(P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\)( 1 )

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow\)\(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)

\(\Leftrightarrow P\ge\dfrac{3}{2}\)

Vậy \(P_{min}=\dfrac{3}{2}\)

Dấu " = " xảy ra khi \(a=b=c\)

x + y = 1

=> (x + y)³ = 1

<=> x³ + y³ + 3x²y + 3xy² = 1

<=> x³ + y³ + 3xy (x+y) = 1

<=> x³ + y³ + 3xy = 1

Vậy ... = 1

x - y = 1

=> (x - y)³ = 1

<=> x³ - y³ - 3x²y + 3xy² = 1

<=> x³ - y³ - 3xy (x - y) = 1

<=> x³ - y³ - 3xy = 1

Vậy ... = 1

a) Vì x + y = 1 => ( x + y )³ = 1

=> x³ + 3x²y + 3xy² + y³ = 1

=> x³ + y³ + 3xy ( x + y ) = 1

=> x³ + y³ +3xy = 1 (do x+y=1)

b) x-y=1 => (x-y)³=1

=> x³ - 3x²y + 3xy² -y³ = 1

=> x³ -y³ - 3xy (x - y) = 1 

=> x³ - y³ -3xy =1 (do x-y=1) 

1)

\(A=\left(x-y+1\right)^2+\left(y-2\right)^2+5\ge5\)

GTNN A=5 khi y=2 và x=1

2)

\(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

\(A=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)

(12x - 5)(4x - 1) + (3x - 7)(1 - 16x) = 81

48x² - 12x - 20x + 5 + 3x - 48x² - 7 + 112x = 81

83x - 2 = 81

83x = 81 + 2

83x = 83

x = 1

nhân ra rồi làm thôi dễ lắm

a)

a) a³ + b³ 

= (a + b)(a² - ab + b²)

= (a + b)(a² + 2ab + b² - 3ab) 
= (a + b)[(a + b)² - 3ab] = (a + b)³ - 3ab(a + b) 

b)

(a - b)³ + 3ab(a - b)

= a³ - 3a².b + 3.ab² - b³+ 3a²b - 3ab²

 = a³- b³

áp dụng

\(a^3+b^3\) 

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=\left(-5\right)^3-3.6.\left(-5\right)\)

\(=-125+90\)

\(=-35\)

này phải nói là bằng 1/2BK hoặc 1/2AC sao lại bằng cạnh đó sai rồi

trên tia đối CA lấy N sao cho CA = CN  

suy ra MC là đường trung bình của tam giác  ABN   

suy ra MC = 1/ 2 BN ( 1 ) 

Xét tam giác ACK và tam giác ABN có : 

AB = AC ( gt ) ; AK = AN ( AB = AC ) ;góc A là góc chung .

Suy ra tam giác ACK = tam giác ABN ( C. G .C )

Suy ra BN = CK ( 2 ) 

TỪ (1) ; (2) Suy ra 1/2 CK = MC .

 kb với mình nha

\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)

\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)

\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)

\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

câu 1 :Phân tích được

(x-3-y).(x-3-y)

2. x² - x - 12 = 0

=> x² - 4x + 3x - 12 = 0

=> x( x - 4 ) + 3( x - 4 ) = 0

=> ( x + 3 )(x - 4 ) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0 \end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

Vậy x=-3; x=4

Ta có :

\(A=3^{4\left(n+1\right)}-4^{3\left(n+1\right)}=81^{n+1}-64^{n+1}\)

\(=\left(81-64\right)\left(81^n+81^{n-1}.64+...+81.64^{n-1}+64^n\right)\)

\(=17\left(81^n+81^{n-1}.64+...+81.64^{n-1}+64^n\right)\)chia hết cho 17

Vậy ...

Ta có n³ - n=n( n²-1)=(n-1)n(n+1)

Mà tích ba số tự nhiên liên tiếp nên chia hết cho 2 và 3 => chia hết cho 6

16(x-y)²-25(x+y)²

=16(x-y)²-25(x-y)²

=(16-25)(x-y)²

=-9(x-y)²

\(=\left[4x-4y\right]^2-\left(5x+5y\right)^2\)

\(=\left(4x-4y+5x-5y\right)\left(4x-4y-5x+5y\right)\)

\(=\left(9x-9y\right)\left(y-x\right)\)

\(=-9\left(x-y\right)^2\)