i: =20*(-29)+(-20)*111

=20(-29-111)

=-20*140=-2800

k: \(=-152\cdot125+152\cdot333-125\cdot333+125\cdot152\)

=27*333

=8991

m: \(=63\left(83+17\right)=6300\)

n: \(=126-7\cdot\left(-16\right)=126+112=238\)

p: \(=35\left(71\cdot2-7-13\right)=35\cdot122=4270\)

q: \(=18\cdot23-18\cdot17-13\cdot23-13\cdot18=5\cdot23-18\cdot30=115-540=-425\)

Bài `4`

`1, 2y(x+2)-3x-6`

`=2y(x+2) -(3x+6)`

`=2y(x+2) -3(x+2)`

`=(x+2)(2y-3)`

`2, 3(x+4) -x^2-4x`

`=3(x+4)-(x^2+4x)`

`=3(x+4) -x(x+4)`

`=(x+3)(3-x)`

`3, 2(x+5) -x^2-5x`

`=2(x+5)-(x^2+5x)`

`=2(x+5)-x(x+5)`

`=(x+5)(2-x)`

`4, x^2 +6x-3(x+6)`

`= (x^2+6x) -3(x+6)`

`=x(x+6)-3(x+6)`

`=(x+6)(x-3)`

`5, x(x+y) -5x-5y`

`=x(x+y) -(5x+5y)`

`=x(x+y)-5(x+y)`

`=(x+y)(x-5)`

`6,x(x-y)+2x-2y`

`=x(x-y)+2(x-y)`

`=(x-y)(x+2`

Bạn tách ra từng bài đi ạ. Làm all trong 1 câu nhiều lắm.

Em ko muốn bài giải vậy Thầy xóa đây. Em tự xem lại cách học của mình nhé!

( x - 1)³ - (x + 3) . (x² - 3x + 9) + 3 .  (x + 2) - (x - 2)  = 2

=>x³-3x².(-1)+3x.(-1)²-(-1)³-x(x²-3x+9)-3(x²-3x+9)+3x+6-x+2=2

x³+3x²+3x+1-x³+3x²-9x-3x²+9x-27+3x+6-x+2=2

(x³-x³)+(3x²+3x²-3x²)+(3x-9x+9x+3x-x)+(1-27+6+2)=2

3x²-5x-18=2

x(3x-5)=20

Thử lần lượt nha bạn

Bài 2

(x+y+z)²-2(x+y+z)(x+y)+(x+y)²

=(x+y+z)²-2x²-4xy-2xz-2yz+x²+2.xy+y²

=z²+(y+x)2z+y²+2xy+x²-2x²-4xy-2z(x+y)+x²+2xy+y²

=z²+(x+y)2z-2z(x+y)+(y²+y²)+(2xy+2xy-4xy)+(x²-2x²+x²)

=z²+2y²

2) nH2=0,1(mol)

a) PTHH: Fe +2  HCl -> FeCl2 + H2

0,1______0,2______0,1____0,1(mol)

nFe=nH2=0,1(mol)

=>mFe=nFe.M(Fe)=0,1.56=5,6(g)

=> mFeO=mX-mFe= 9,2-5,6=3,6(g)

=> nFeO=mFeO/M(FeO)=3,6/72=0,05(mol)

PTHH: FeO +2 HCl -> FeCl2 + H2

0,05_________0,1___0,05__0,05(mol)

b) Sao lại mỗi oxit a, có một oxit thôi mà :(  Chắc % KL mỗi chất.

%mFeO=(mFeO/mhh).100%=(3,6/9,2).100=39,13%

=>%mFe=100%-%mFeO=100%-39,13%=60,87%

c) nHCl(tổng)= 2.nFe +2.nFeO=2.0,1+2.0,05=0,3(mol)

=>mHCl=nHCl.M(HCl)=0,3.36,5=10,95(g)

=>mddHCl=(mHCl.100%/C%ddHCl=(10,95.100)/7,3=150(g)

d) - Dung dich thu được chứa FeCl2.

mFeCl2=nFeCl2(tổng) . M(FeCl2)= (0,1+0,05).127=19,05(g)

mddFeCl2=mddHCl+mhh-mH2=150+9,2-0,1.2=159(g)

=> C%ddFeCl2=(mFeCl2/mddFeCl2).100%=(19,05/159).100=11,981%

`a,-2022.(x+8)=0`

`=>x+8=0:(-2022)`

`=>x+8=0`

`=>x=0-8`

`=>x=-8`

`b,(7-x)(x+3)=0`

`@ TH1`

`7-x=0`

`=>x=7-0`

`=>x=7`

`@ TH2`

`x+3=0`

`=>x=0-3`

`=>x=-3`

`c,2023x .(14-x)=0`

`@ TH1`

`2023x=0`

`=>x=0:2023`

`=>x=0`

`@ TH2`

`14-x=0`

`=>x=14-0`

`=>x=14`

`d,x^2-x=0`

`=>x(x-1)=0`

`@ TH1`

`x=0`

`@ TH2`

`x-1=0`

`=>x=0+1`

`=>x=1`

\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\left(-2020\right).\left(x+8\right)=0\)

\(=>x+8=0=>x=-8\)

\(\left(7-x\right).\left(x+3\right)=0\)

\(=>\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(2023x.\left(14-x\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\14-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)

\(x^2-x=0\)

\(x\left(x-1\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

            Bài làm :

Áp dụng tính chất của dãy tỉ số bằng nhau ; ta có :

\(\frac{x}{5}=\frac{y}{6}=\frac{x+y}{5+6}=\frac{44}{11}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=4\\\frac{y}{6}=4\end{cases}\Rightarrow}\hept{\begin{cases}x=20\\y=24\end{cases}}\)

Mà ta có :

\(\frac{y}{8}=\frac{z}{11}\Rightarrow\frac{24}{8}=\frac{z}{11}\Rightarrow z=33\)

Vậy :

\(A=x-y-2z=20-24-2\times33=-70\)

\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{11}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{40}=\frac{y}{48}\\\frac{y}{48}=\frac{z}{66}\end{cases}}\Leftrightarrow\frac{x}{40}=\frac{y}{48}=\frac{z}{66}\)

Theo tính chaasts dãy tỉ số bằng nhau

\(\frac{x}{40}=\frac{y}{48}=\frac{z}{66}=\frac{x+y-z}{40+48-66}=\frac{44}{22}=2\)

\(\hept{\begin{cases}x=40.2=80\\y=48.2=96\\z=66.2=132\end{cases}}\)

Ta có \(A=x-y-2z\Leftrightarrow A=80-96-2.132=-280\)

Vậy giá trị biểu thức A là -280

Bỏ () Rồi tính nhé