\(\sqrt{x^2-2x+4}+\sqrt{x^2+5}=9-2x\left(đk:x\le\dfrac{9}{2}\right)\)

\(\Leftrightarrow x^2-2x+4+x^2+5+2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=81-36x+4x^2\)

\(\Leftrightarrow2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=2x^2-34x+72\)

\(\Leftrightarrow4\left(x^2-2x+4\right)\left(x^2+5\right)=4x^4+1156x^2+5184-136x^3+288x^2-4896x\)

\(\Leftrightarrow4x^4-8x^3+36x^2-40x+80=4x^4-136x^3+1444x^2-4896x+5184\)

\(\Leftrightarrow128x^3-1408x^2+4856x-5104=0\)

\(\Leftrightarrow128x^2\left(x-2\right)-1152x\left(x-2\right)+2552\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(128x^2-1152x+2552\right)=0\)

\(\Leftrightarrow x=2\left(tm\right)\)(do \(128x^2-1152x+2552>0\))

a, 2/3 - 2/9 + 7/9 = 6/9 - 2/9 + 7/9 = 10/9
b, 7/6 + 3/5 : 6 = 7/6 + 3/5 x 1/6 = 7/6 + 1/10 = 70/60 + 6/10 = 76/10 = 38/5
c, x : 6/25 = 18

x = 18 x 6/25

x = 36/25
d, ( 2/5 + 4/7) : x = 17/5

29/35 : x = 17/5

x = 29/35 : 17/5

x = 29/119

\(a.\dfrac{2}{3}-\dfrac{2}{9}+\dfrac{7}{9}=\dfrac{6}{9}-\dfrac{2}{9}+\dfrac{7}{9}=\dfrac{4}{9}+\dfrac{7}{9}=\dfrac{11}{9}\\ b.\dfrac{7}{6}+\dfrac{3}{5}:6=\dfrac{7}{6}+\dfrac{3}{5}\times\dfrac{1}{6}=\dfrac{7}{6}+\dfrac{1}{10}=\dfrac{70}{60}+\dfrac{6}{60}=\dfrac{76}{60}=\dfrac{19}{15}\\ c.x:\dfrac{6}{25}=18\\ x=18\times\dfrac{6}{25}\\ x=\dfrac{108}{25}\\ d.\left(\dfrac{2}{5}+\dfrac{4}{7}\right):x=\dfrac{17}{5}\\ \dfrac{34}{35}:x=\dfrac{7}{5}\\ x=\dfrac{34}{35}:\dfrac{7}{5}\\ x=\dfrac{34}{49}\)

x - 18 - 42 = 23 - 43 - 70 - x

\(\rightarrow\) 2x = -30

\(\rightarrow\) x = -15

\(\left(-\left(x+15\right)\right)^2\) - 19 = \(3^2\) .5

\(\rightarrow\)\(x^2+30x+225\) - 19 = 45

\(\rightarrow x^2+30x+161=0\)

\(\rightarrow x^2+23x+7x+161=0\)

\(\rightarrow x\left(x+23\right)+7\left(x+23\right)=0\)

\(\rightarrow\left(x+7\right)\left(x+23\right)=0\)

\(\rightarrow\) x = -7 hoặc x = -23

<=> 2x + 3 - 8 + 2x = 5 hoặc -2x - 3 + 8 - 2x = 5 

<=> 4x - 5 = 5 hoặc -4x + 5 = 5 

<=> 4x = 10 hoặc -4x = 0 

<=> x = 5 / 2 hoặc x = 0   

Ta có \(\left(x+y\right)^2=x^2+2xy+y^2=49\Leftrightarrow xy=\dfrac{49-25}{2}=12\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=25^2-2\cdot12^2=337\)

Ta có \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=7^3-3\cdot12\cdot7=91\)

\(\left(x^2+y^2\right)\left(x^3+y^3\right)=91\cdot25=2275\\ \Leftrightarrow x^5+y^5+2x^2y^2\left(x+y\right)=2275\\ \Leftrightarrow x^5+y^5=2275-2\cdot144\cdot7=259\)

3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5

       1/3 x (2,48+0,52) x y : 60 : 5 = 2/5

    1/3 x 3 x y : 60 : 5                     = 2/5

                      y: 60 : 5                    =2/5

                      y: 60                          =2/5 x 5

                      y : 60                         =2 

                      y                                =  2 x 60

                      y                                  =120

Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)

hay y=120

Do \(\left|x\right|,\left|x^2+x\right|\ge0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2+x=0\end{matrix}\right.\)

\(\Rightarrow x=0\)

x = 0

Vì bất cứ số nào nhân hay chia với o đều = 0

77/45

7/6

29/10

\(\dfrac{3}{5}+\dfrac{4}{3}\times\dfrac{5}{6}=\dfrac{3}{5}+\dfrac{10}{9}=\dfrac{27}{45}+\dfrac{50}{45}=\dfrac{77}{45}\)

\(2-\dfrac{7}{6}+\dfrac{1}{3}=\dfrac{12}{6}-\dfrac{7}{6}+\dfrac{2}{6}=\dfrac{7}{6}\)

\(\dfrac{7}{6}\times3-\dfrac{3}{5}=\dfrac{7}{2}-\dfrac{3}{5}=\dfrac{35}{10}-\dfrac{6}{10}=\dfrac{29}{10}\)

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