a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn

Câu 1:

a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$

b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)

c.

\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)

d.

\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)

Câu 2:

a. $x\times \frac{2}{7}=\frac{6}{11}$

$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$

b. $x: \frac{3}{2}=\frac{1}{4}$

$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$

Câu 3:

Coi số lớn là 5 phần thì số bé là 3 phần. Hiệu số phần bằng nhau là:
$5-3=2$ (phần)

Số lớn là: $36:2\times 5=90$

Số bé là: $36:2\times 3=54$

Câu 2: A

Câu 1: B

Câu 13 :

\(\left(-\frac{1}{4}+\frac{5}{8}\right)+-\frac{3}{5}\)

\(=\frac{3}{8}-\frac{-3}{5}\)

\(=\frac{39}{40}\)

Câu 14 :

\(M=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)

\(=\frac{5}{9}.1=\frac{5}{9}\)

Câu 15 :

\(E=\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=0\)

Câu 16 :

\(H=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}+\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\frac{-7}{24}=-3\)

huhuhuhuhuhuhuhuhuhuhuhuhu giúp mk đi 

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Câu 14:

M =  \(\frac{5}{9}\) \(.\) \(\frac{7}{13}\) \(+\) \(\frac{5}{9}\) \(.\) \(\frac{9}{13}\) \(-\) \(\frac{5}{9}\) \(.\) \(\frac{3}{13}\)

M =    \(\frac{5}{9}\)  \(.\) (   \(\frac{7}{13}\)\(+\) \(\frac{9}{13}\) \(-\) \(\frac{3}{13}\))

M =      \(\frac{5}{9}\)  \(.\)    \(1\)

M  =              \(\frac{5}{9}\)

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

Câu 1 : x+1=1 => x = 0 => pt trên =-1 loại

              x+1 = 3 => x= 2 => 2y-1=3 => y=2

vậy x=2;y=2

câu 2 : 2x-1 = 1 > x = 1 ;  y +4=7 => y=3

           2x-1 = 7 => x=4 ; y +7 = 1 => y = -6 loại

vậy x=1, y=3 v

Câu 4  tương tự x=0 y=3

a: =7/8:(2/9-18+1/36)-5/12

=-7/142-5/12=-397/852

b: =3/7(4/9+5/9:6/12)=2/3

c: =5^8(16/31-47/31)+1/3=-5^8+1/3

d: =7/2(3/8+5/8:4/15)=609/64