`a)`

Ta có: \(PTK_{Ba\left(NO_3\right)_y}=261\) \((g/mol)\)

\(\Leftrightarrow137+\left(14+16.3\right).y=261\)

\(\Leftrightarrow y=2\)

`=>` CTHH: Ba(NO₃)₂

`=>` Hóa trị của nhóm NO₃ là I

`b)`Ta có: \(PTK_{Mn_2O_x}=222\) \((g/mol)\)

\(\Leftrightarrow55.2+16.x=222\)

`<=>x=7`

`=>` CTHH: Mn₂O₇

`=>` Hóa trị của Mn là VII

Công thức hóa học: \(Mg\left(HCO_3\right)_2\)

PTK: \(24+2.\left(1+12+16.3\right)=146\) \((g/mol)\)

\(4x^2+6x+7+\sqrt{x^2+3x+9}=15\)

Đặt \(2x^2+3x=t\)

Ptr trở thành:

\(2t+7+\sqrt{t+9}=15\)

\(\Leftrightarrow\sqrt{t+9}=8-2t\)

\(\Leftrightarrow t+9=64-32t+4t^2\)

\(\Leftrightarrow4t^2-33t+55=0\)

\(\Delta=\left(-33\right)^2-4.4.55=209>0\)

\(\rightarrow\left\{{}\begin{matrix}t=\dfrac{33+\sqrt{209}}{8}\\t=\dfrac{33-\sqrt{209}}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-\dfrac{33+\sqrt{209}}{8}=0\\2x^2+3x-\dfrac{33-\sqrt{209}}{8}=0\end{matrix}\right.\) `=>` Casio

Gọi `x`\(\left(x\ge0\right)\) là số gam nước thêm vào `=>` Số gam nước lúc sau là: `300+x` (g)

Dd chứa 15% muối, ta có pt:

\(\dfrac{75}{300+x}.100=15\)

\(\Leftrightarrow x=200\left(tm\right)\)

Vậy cần đổ thêm 200 gam nước để được dd 15% muối

\(x^3-6x^2+12x-8=0\)

`<=>x^3-3.x^3.2+3.x.2^2-2^3=0`

`<=>(x-2)^3=0`

`<=>x-2=0`

`<=>x=2`

\(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)

\(\Leftrightarrow\left(x+3\right)^3=0\)

`<=>x+3=0`

`<=>x=-3`

`a)`\(\widehat{C}+\widehat{B}=90^o\) \(\Rightarrow\widehat{A}=90^o\)

Ta có: \(\left\{{}\begin{matrix}\widehat{ABH}+\widehat{BAH}=90^o\\\widehat{CAH}+\widehat{BAH}=90^o\end{matrix}\right.\) \(\Rightarrow\widehat{ABH}=\widehat{CAH}\)

Xét tam giác ABH và tam giác CAH, có:

\(\widehat{AHB}=\widehat{AHC}=90^o\left(gt\right)\)

\(\widehat{ABH}=\widehat{CAH}\left(cmt\right)\)

Vậy tam giác ABH đồng dạng tam giác CAH ( g.g )

\(\Rightarrow\dfrac{AH}{CH}=\dfrac{BH}{AH}\)

\(\Leftrightarrow AH^2=BH.CH\) 

`@`\(\left(x^2+x\right)^2-6x^2-6x+8=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-4\left(x^2+x\right)+8=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-4\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\); \(\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{17}}{2}\\x=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\)

`@`\(\left(x^2+10x+16\right)\left(x+4\right)\left(x+6\right)-20=0\) 

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)-20=0\)

\(\Leftrightarrow\left(x^2+10x\right)^2+24\left(x^2+10x\right)+16\left(x^2+10x\right)+384-20=0\)

\(\Leftrightarrow\left(x^2+10x\right)^2+40\left(x^2+10x\right)+364=0\)

\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+14\right)+26\left(x^2+10x+14\right)=0\)

\(\Leftrightarrow\left(x^2+10x+14\right)\left(x^2+10x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5+\sqrt{11}\\x=-5-\sqrt{11}\end{matrix}\right.\)

chịu òi:v ko ra =((