\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

Bài 19:

a: \(A=5x+\dfrac{1}{9}y=5\cdot\dfrac{-1}{10}+\dfrac{1}{9}\cdot4.8=\dfrac{-1}{2}+\dfrac{8}{15}=\dfrac{-15+16}{30}=\dfrac{1}{30}\)

b: \(A=x-\dfrac{2}{3}=\dfrac{-1}{3}-\dfrac{2}{3}=-1\)

\(a,7x-2x-\dfrac{2}{3}y+\dfrac{7}{9}y=5x+\dfrac{1}{9}y\\ =5.\left(\dfrac{-1}{10}\right)+\dfrac{1}{9}.4,8\\ =\dfrac{-1}{2}+\dfrac{8}{15}=\dfrac{1}{30}\\ b,x=\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\\ =\dfrac{-1}{3}+\dfrac{\dfrac{-7}{40}+\dfrac{5}{11}}{\dfrac{21}{80}-\dfrac{15}{22}}\\ =\dfrac{-1}{3}+\dfrac{\dfrac{123}{440}}{\dfrac{-369}{880}}=\dfrac{-1}{3}+\dfrac{-2}{3}=\dfrac{-3}{3}=\left(-1\right)\)

có bài 2 nào đâu

Bài 4 ý

Bài 5 ý

12:

\(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\dfrac{\sqrt{y}}{y}\cdot\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{y}}\)

a) \(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}\)

\(=\sqrt{\dfrac{x}{y}}\)

Em hãy đăng bài ở môn Toán nhé!

Bài 2: 

a: \(8x^3-36x^2+54x-27=\left(2x-3\right)^3\)

b: \(x^3+6x^2+12x+8=\left(x+2\right)^3\)

c: \(27x^3+8y^3=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)

d: \(x^3-\dfrac{y^3}{8}=\left(x-\dfrac{1}{2}y\right)\left(x^2+\dfrac{1}{2}xy+\dfrac{1}{4}y^2\right)\)

2.1 

\(\Leftrightarrow x^3+3x^2+2x-3x^2-9x-6=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

2.2

\(\Leftrightarrow x^3-2x^2-2x-x^2+2x+2=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\pm\sqrt{3}\end{matrix}\right.\)

2.3

\(\Leftrightarrow3x^3-3x^2+2x+3x^2-3x+2=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x^2-3x+2\right)=0\)

\(\Leftrightarrow x=-1\)

2.5

\(\Leftrightarrow2x^3+x^2+3x-4x^2-2x-6=0\)

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

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

\(\Leftrightarrow x=2\)

2.9

\(\Leftrightarrow3x^3-3x^2-3x=1\)

\(\Leftrightarrow3x^3=3x^2+3x+1\)

\(\Leftrightarrow4x^3=x^3+3x^2+3x+1\)

\(\Leftrightarrow4x^3=\left(x+1\right)^3\)

\(\Leftrightarrow\left(x\sqrt[3]{4}\right)^3=\left(x+1\right)^3\)

\(\Leftrightarrow x\sqrt[3]{4}=x+1\)

\(\Leftrightarrow x\left(\sqrt[3]{4}-1\right)=1\)

\(\Leftrightarrow x=\dfrac{1}{\sqrt[3]{4}-1}\)

cái gì vậy bạn

? bài ở đâu

ko đăng ảnh đc ạ 

a) Gọi x, y lần lượt là số mol Al, Fe

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Fe + H2SO4 → FeSO4+ H2

\(\left\{{}\begin{matrix}27x+56y=5,54\\\dfrac{3}{2}x+y=\dfrac{3,584}{22,4}\end{matrix}\right.\)

=> x=0,06 , y =0,07

=> \(m_{Al}=1,62\left(g\right);m_{Fe}=3,92\left(g\right)\)

b) \(n_{H_2SO_4\left(pứ\right)}=n_{H_2}=0,16\left(mol\right)\)

=> \(m_{H_2SO_4\left(pứ\right)}=0,16.98=15,68\left(g\right)\)

c)  \(m_{ddH_2SO_4}=\dfrac{15,68}{20\%}=78,4\left(g\right)\)

c) 2NaOH + H2SO4 → Na2SO4 + 2H2O

\(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,25.0,6=0,075\left(mol\right)\)

=> \(m_{H_2SO_4\left(bđ\right)}=15,68+0,075.98=23,03\left(g\right)\)

4:

d: