(a +b – 1) x (a +b -2 ) x.. x… x (a +b -100)

=(49 +51 – 1) x (49 +51 -2 ) x.. x… x (49 +51 -100)

=(49 +51 – 1) x (49 +51 -2 ) x.. x… x (100-100)

=(49 +51 – 1) x (49 +51 -2 ) x.. x… x 0 =0

๛нỌ¢ тỐтツ nha em

Tham lam

Ko chép đề

\(2)x-17+x=x-7\)

\(x+x-x=-7+17\)

\(x=-10\)

Vậy ....

âm 10 , nhỉ ?

a: \(=68\cdot72\cdot49\cdot\left[324-324\right]=0\)

b: \(=\left(1250-1250\right)\cdot\left(1+...+100\right)=0\)

\(a,\left(68\times72\times49\right)\times\left(27\times12-9\times36\right)=\left(68\times72\times49\right)\times0=0\\ b,\left(25\times50-125\times10\right)\times\left(1+2+...+100\right)=0\times\left(1+2+...+100\right)=0\)

a, ( 68 × 72 × 49 ) × ( 27 × 12 − 9 × 36 ) = ( 68 × 72 × 49 ) × 0 =

b, ( 25 × 50 − 125 × 10 ) × ( 1 + 2 + ... + 100 ) = 0 × ( 1 + 2 + ... + 100 ) = 0

2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)

\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)

\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)

3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)

Câu 1 :

\(a,\left(3x+2\right)^2=9x^2+12x+4.\)

\(b,\left(6a^2-b\right)^2=36a^4-12a^2b-b^2\)

\(c,\left(4x-1\right)\left(4x+1\right)=16x^2-1\)

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

\(e,\left(a^2+b^2\right)\left(a^2-b^2\right)=a^4-b^4\)

\(f,\left(x^3+y^2\right)\left(x^3-y^2\right)=x^6-y^4\)

Bài 2 :

\(a,A=9x^2+42x+49=9+42+49=100.\)

\(b,B=25x^2-2xy+\frac{1}{25}y^2=\left(5x^2\right)-2.5x.\frac{1}{5}y+\left(\frac{1}{5}y\right)^2\)

\(=\left(5x-\frac{1}{5}y\right)^2=\left(-1+1\right)^2=0\)

\(c,C=4x^2-28x+49=4x^2-14x-14x+49\)

\(=2x\left(x-7\right)-7\left(x-7\right)=\left(2x-7\right)\left(x-7\right)\)

\(=\left(8-7\right)\left(4-7\right)=-3\)

Bài 3 :

\(a,\left(x-3\right)^2-4=0\)

\(\Rightarrow\left(x-3+2\right)\left(x-3-2\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Rightarrow x\in\left\{1;5\right\}\)

\(b,x^2-2x=24\)

\(\Rightarrow x^2-2x-24=0\)

\(\Rightarrow x^2+4x-6x-24=0\)

\(\Rightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{-4;6\right\}\)

câu 4 a=12 b=13

Bài 1 : x/3 = y/4 = z/5 => x²/9 = y²/16 = z²/25 
=> 2x²/18 = 2y²/32 = 3z²/75 
=> x²/9 = (2x² + 2y² - 3z²)/(18 + 32 - 75) = - 100/(-25) = 1/4 
=> x²/9 = 1/4 => x² = 9/4 => x = ±3/2 
y²/16 = 1/4 => y² = 4 => y = ± 2 
z²/25 = 1/4 => z² = 25/4 => z = ±5/2 
Mà x, y, z cùng dấu. 
Vậy (x ; y ; z) = (3/2 ; 2 ; 5/2) , (-3/2 ; -2 ; -5/2)

B3 ko tìm được x,y,z thỏa mãn do kết quả là 1 số không dương