|x+8/5| + |2,2-2y| = 0 ( không thể < 0 )

=> x + 8/5 = 2,2 - 2y = 0

=> x = -8/5; 2y = 2,2

=> x = -8/5; y = 1,1

\(y=1,1\)

x=-8/5

Câu a:

x^2 - x + 1 = 0

x^2 - 2.x.1/2 + 1/4 + 3/4 = 0

(x - 1/2)^2 + 3/4 = 0

Vì (x - 1/2)^2 ≥ 0 ∀ x;

(x - 1/2)^2 + 3/4 ≥ 3/4 > 0 ∀ x

Vậy x ∈ ∅

Câu b:

x^2 + 2xy+ 2y^2 + 2y + 1 = 0

(x^2 + 2xy + y^2) + (y^2 + 2y+ 1) = 0

(x + y)^2 + (y + 1)^2 = 0 (1)

Vì (x+ y)^2 ≥ 0 ∀ x; y (y + 1)^2 ≥ 0 ∀ y nên (1) xảy ra khi và chỉ khi:

x + y = 0 và y +1 = 0

y+ 1 = 0; y = -1

x + y = 0

x - 1 = 0

x = 1

Vậy (x; y)= (1; -1)

\(A=\left(13+x\right)\left(17+x\right)\left(2-x\right)\le0\)

Nếu  \(x< -17\), ta có 13 + x < 0, 17 + x \(\le\) 0, 2 - x > 0 

Vậy nên A \(>\) 0,

Nếu  \(-17\le x\le-13\),  ta có: 13 + x < 0 , 17 + x > 0, 12 - x > 0. Vậy thì \(A\le0\)

Nếu  \(-13< x< 2\), ta có: 13 + x > 0, 17 + x > 0, 2 - x > 0. Vậy nên \(A>0\)

Nếu \(x\ge2\) , ta có \(13+x>0,17+x>0,2-x\ge0\). Vậy nên \(A\le0\)

Vậy để \(A\le0\) thì \(-17\le x\le-13\) hoặc \(x\ge2.\)

trả lời được tớ cho3 k

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