`Answer:`

`x+9/4 = 1/3-2/4 - (-1/5)- (-2/3) -(-3/4)+4/5`

`<=>x+9/4=1/3-2/4+1/5+2/3+3/4+4/5`

`<=>x=1/3-1/2+1/5+2/3+3/4+4/5-9/4`

`<=>x=(1/2+2/3)+(3/4-9/4)+(1/5+4/5)-1/2`

`<=>1-3/2+1-1/2`

`<=>2-2=0`

Vậy `x=0`

\(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\)

\(\Leftrightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\)

\(\Leftrightarrow\frac{17}{3}x=\frac{9}{2}-\frac{5}{3}\)

\(\Leftrightarrow\frac{17}{3}x=\frac{17}{6}\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy..

Học tốt

#Thảo Vy#

Câu a:

(x^2 + 1)^2 + (y^2 - 4)^2 = 0

Vì (x^2 + 1)^2 ≥ 0 ∀ x; (y^2 - 4)^2 ≥ 0 ∀ y nên:

x^2 + 1 = 0 và y^2 - 4 = 0

x^2 = - 1 (vô lí)

Vậy không có giá trị nào của x; y thỏa mãn đề bài;

Hay x; y ∈ ∅

Câu b:

b; (x^2 - 9^2)^2 + (y^2 - 16)^4 = 0 (1)

Vì (x^2 - 9^2)^2 ≥ 0 ∀ x; (y^2 - 16)^4 ≥ 0 ∀ y nên (1) xảy ra khi

x^2 - 9^2 = 0; y^2 - 16 = 0

x^2 = 9^2

x = -9 hoặc x = 9

y^2 = 16

y = -4 hoặc y = 4

Vậy (x; y) = (9; -4); (9; -4); (-9; 4); (-9; 4)

D.2

cà khịa time x=+
 

cân cặc

1 - 8x⁷ - x² + 5x⁴ - 2 + 3x⁵ + 8x⁷

= 3x⁵ + 5x⁴ - x² - 1

1 - 8x7 - x2 + 5x4 - 2 + 3x5 + 8x7

= 3x5 + 5x4 - x2 - 1

a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

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