đề?

\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)

Đặt a = \(\sqrt{x^2+2x-1}\left(a\ge0\right)\) , ta đc pt: 2(1 - x).a = a² - 4x => a² - 2(1 - x)a - 4x = 0

Ta có: \(\Delta'=\left[-\left(1-x\right)\right]^2+4x=1-2x+x^2+4x=x^2+2x+1=\left(x+1\right)^2\)\(\Rightarrow\sqrt{\Delta'}=x+1\)

\(\Rightarrow\left[\begin{array}{nghiempt}a=\frac{1-x+x+1}{1}=2\\a=\frac{1-x-x-1}{1}=-2x\left(vn\right)\end{array}\right.\)

+) Với a = 2 \(\Rightarrow\sqrt{x^2+2x-1}=2\Rightarrow x^2+2x-1=4\Rightarrow x^2+2x-5=0\Rightarrow\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)

                                 Vậy pt có 2 nghiệm \(\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)

ĐK:...

\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)

\(\Leftrightarrow2\left(1-x\right)\sqrt{\left(1+x\right)^2-2}=\left(1-x\right)^2-2\)

Đặt \(\begin{cases}a=1+x\\b=1-x\end{cases}\),ta có hệ:

\(\begin{cases}2b\sqrt{a^2-2}=b^2-2\\a+b=2\end{cases}\)

\(\Leftrightarrow\begin{cases}4a^2b^2-8b^2=b^4-4b^2+4\\a+b=2\end{cases}\)

\(\Leftrightarrow\begin{cases}4a^2b^2=b^4+4b^2+4\\a+b=2\end{cases}\)

\(\Leftrightarrow\begin{cases}2ab=b^2+2\\b=2-a\end{cases}\)hay\(\begin{cases}2ab=-b^2-2\\b=2-a\end{cases}\)

\(\Leftrightarrow2a\left(2-a\right)=\left(2-a\right)^2+2\)hay\(2a\left(2-a\right)=-\left(2-a\right)^2-2\)

\(\Leftrightarrow3a^2-8a+6=0\)hay a²=6

\(\Rightarrow\left[\begin{array}{nghiempt}a=x+1=\sqrt{6}\\a=x+1=-\sqrt{6}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{array}\right.\)

ê ku vĩ căn đen ta =x+1 thế GTTĐ đâu quên cơ bản ak

Câu 3:

\(a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ b,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{Fe}=n_{H_2}=0,3(mol)\\ \Rightarrow m_{Fe}=0,3.56=16,8(g)\\ \Rightarrow m_{Fe_2O_3}=32,8-16,8=16(g)\\\)

\(c,V_{dd_{H_2SO_4}}=\dfrac{294}{1,2}=245(ml)\\ n_{FeSO_4}=n_{Fe}=0,3(mol)\\ n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=\dfrac{16}{160}=0,1(mol)\\ \Rightarrow C_{M_{FeSO_4}}=\dfrac{0,1}{0,245}=0,41M\\ C_{M_{Fe_2(SO_4)_3}}=\dfrac{0,3}{0,245}=1,22M\)

Câu 1:

\(BaCO_3\xrightarrow[]{t^o}BaO+CO_2\uparrow\\ BaO+H_2O\longrightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+SO_2\longrightarrow BaSO_3+H_2O\\ BaSO_3+2HCl\longrightarrow BaCl_2+SO_2\uparrow+H_2O\)

Câu 2:

- Trích mẫu thử, cho quỳ tím vào từng mẫu thử:

+ Quỳ hóa đỏ: \(H_2SO_4\)

+ Quỳ hóa xanh: \(NaOH\)

+ Quỳ ko đổi màu: \(NaCl;Na_2SO_4(1)\)

- Cho \(BaCl_2\) vào nhóm \((1)\):

+ Xuất hiện kết tủa trắng: \(Na_2SO_4\)

+ Ko hiện tượng: \(NaCl\)

\(PTHH:Na_2SO_4+BaCl_2\to BaSO_4\downarrow+2NaCl\)

17)\(\left(x+y+z\right)^2-4z^2\)

\(=\left(x+y+z-2z\right)\left(x+y+z+2z\right)\)

\(=\left(x+y-z\right)\left(x+y+3z\right)\)

18)\(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)

19)\(8x^3-y^3-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2-6xy\right)\)

\(=\left(2x-y\right)\left(4x^2-4xy+y^2\right)\)

\(=\left(2x-y\right)\left(2x-y\right)^2=\left(2x-y\right)^3\)

20)\(-\frac{1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2\)

\(=-\left(\frac{1}{9}x^2-\frac{1}{3}xy+\frac{1}{4}y^2\right)=-\left(\frac{1}{3}x-\frac{1}{2}y\right)^2\)

21)\(x^4y^4-z^4=\left[\left(xy\right)^2\right]^2-\left(z^2\right)^2\)

\(=\left(x^2y^2-z^2\right)\left(x^2y^2+z^2\right)\)

\(=\left(xy-z\right)\left(xy+z\right)\left(x^2y^2+z^2\right)\)

mơ đi cưng 

tự nghĩ ok

bạn gửi luôn đi

bài 1: thực hiện phép tính:

a/\(\left(\dfrac{2}{5}-\dfrac{3}{4}\right):\dfrac{7}{10}=\left(\dfrac{8}{20}-\dfrac{15}{20}\right):\dfrac{7}{10}=\dfrac{-7}{20}:\dfrac{7}{10}=\dfrac{-1}{2}\)

Bài 5.

a. $A=\frac{3n+2}{n-1}$ chứ nhỉ.

Để $A$ nguyên thì $3n+2\vdots n-1$

$\Leftrightarrow 3(n-1)+5\vdots n-1$

$\Leftrightarrow 5\vdots n-1$

$\Rightarrow n-1\in$ Ư(5)$

$\Rightarrow n-1\in\left\{\pm 1;\pm 5\right\}$

$\Rightarrow n\in\left\{0;2;-4;6\right\}$

b.

$M=\frac{9}{2}\left(\frac{1}{3.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)$

$=\frac{9}{4}\left(\frac{2}{21}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)$

$=\frac{9}{4}\left(\frac{2}{21}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{13}-\frac{1}{15}\right)$
$=\frac{9}{4}\left(\frac{2}{21}+\frac{1}{7}-\frac{1}{15}\right)$

$=\frac{27}{70}$

Giải:

a) Để \(A=\dfrac{3n+2}{n-1}\) là số nguyên thì \(3n+2⋮n-1\) 

\(3n+2⋮n-1\) 

\(\Rightarrow3n-3+5⋮n-1\) 

\(\Rightarrow5⋮n-1\) 

\(\Rightarrow n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\) 

Ta có bảng giá trị:

Vậy \(n\in\left\{-4;0;2;6\right\}\) 
b) \(M=\dfrac{3^2}{3.14}+\dfrac{3^2}{7.18}+\dfrac{3^2}{9.22}+\dfrac{3^2}{11.26}+\dfrac{3^2}{13.30}\) 

\(M=\dfrac{9}{2}.\left(\dfrac{1}{3.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\right)\) 

\(M=\dfrac{9}{2}.\dfrac{1}{2}.\left(\dfrac{2}{21}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\right)\) 

\(M=\dfrac{9}{4}.\left(\dfrac{2}{21}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\)\(M=\dfrac{9}{4}.\left(\dfrac{2}{21}+\dfrac{1}{7}-\dfrac{1}{15}\right)\)

\(M=\dfrac{9}{4}.\dfrac{6}{35}\) 

\(M=\dfrac{27}{70}\)

9)\(x^6+216=\left(x^2\right)^3+6^3=\left(x^2+6\right)\left(x^4+6x^2+36\right)\)

10)\(x^2+12x+36=\left(x+6\right)^2=\left(x+6\right)\left(x+6\right)\)

11)\(9x^2-12xy+4y^2=\left(3x-2y\right)^2=\left(3x-2y\right)\left(3x-2y\right)\)

12)\(-25x^2y^2+10xy-1=-\left(25x^2y^2-10xy+1\right)=-\left(5xy+1\right)^2=-\left(5xy+1\right)\left(5x+1\right)\)

13)\(a^3-6a^2+12a-8=\left(a^3-8\right)-\left(6a^2-12a\right)=\left(a-2\right)\left(a^2+2a+4\right)-6a\left(a-2\right)\)

\(=\left(a-2\right)\left(a^2+2a+4-6a\right)=\left(a-2\right)\left(a^2-4a+4\right)=\left(a-2\right)\left(a-2\right)^2\)