Bạn ơi, ps cuối ah sai rồi nhé^^...

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coi ruộng 1 là 4 phần = nhau thì ruộng 2 là 5 phần = nhau như vậy.

     giá trị 1 phần là :

            4.5 : 9 = 0.5 (tấn thóc)

     tấn 1 :     0.5 x 4 = 2 (tấn)

     tấn 2 :     4.5 - 2 = 2.5(tấn)

thửa ruộng 1:2 tấn thóc

thửa ruộng 2 : 2.5 tấn thóc

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(=>11-3x+1=\frac{9}{2}-5+3,5x\)

\(=>-3x+12=3,5x-\frac{1}{2}\)

\(=>-3x-3,5x=-\frac{1}{2}-12\)

\(=>-6,5x=-12,5\)

\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)

Ủng hộ nha

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(11-3x+1=\frac{9}{2}-5+3,5x\)

\(12-3x=-\left(0,5\right)+3,5x\)

\(12,5-3x=3,5x\)

\(12,5=6,5x\)

\(x=12,5:6,5=\frac{25}{13}\)

-2,5 + |3x + 5| = -1,5

|3x + 5| = -1,5 + 2,5

|3x + 5| = 1

Với x -5/3 ta có:

3x + 5 = 1

3x = 1 - 5

3x = -4

x = -4/3 (nhận)

Với x < -5/3 ta có:

3x + 5 = -1

3x = -1 - 5

3x = -6

x = -6/3

x = -2 (nhận)

Vậy x = -2; x = -4/3

Bài 10 nka m.n

Bài 9:

a) \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\) (ĐK: \(x\ne1;x\ne4;x>0\))

\(P=\left[\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right]\)

\(P=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)

\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)

\(P=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)

\(P=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) 

b) \(P>\dfrac{1}{6}\) khi 

\(\Rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-2\right)-\sqrt{a}}{2\cdot3\sqrt{a}}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}>0\) 

Mà: \(6\sqrt{a}>0\)

\(\Leftrightarrow\sqrt{a}-4>0\)

\(\Leftrightarrow\sqrt{a}>4\) 

\(\Leftrightarrow a>16\)

Vậy: ... 

20:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{1-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\cdot2\)

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

b: \(x+\sqrt{x}+1=\left(\sqrt{x}+1\right)\cdot\sqrt{x}+1>=1>0\)

2>0

Do đó: \(P=\dfrac{2}{x+\sqrt{x}+1}>0\)

8:

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)

\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)

\(=\dfrac{a-1}{4a}\cdot\left(-4\sqrt{a}\right)=-\dfrac{a-1}{\sqrt{a}}\)

b: P<0

=>\(-\left(a-1\right)< 0\)

=>a-1>0

=>a>1

c: P=-2

=>\(\dfrac{a-1}{\sqrt{a}}=2\)

=>\(a-1=2\sqrt{a}\)

=>\(a-2\sqrt{a}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{a}=1+\sqrt{2}\left(nhận\right)\\\sqrt{a}=1-\sqrt{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow a=3+2\sqrt{2}\)

ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)

\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)

Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé

\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)

\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)

\(\Leftrightarrow4x+10=0\)

\(\Leftrightarrow4x=-10\)

\(\Leftrightarrow x=\frac{-5}{2}\)

Trả lời:

( x + 2 )² - 2 ( x + 2 ) ( x + 3 ) + ( x + 5 )² = 7

<=> x² + 4x + 4 - 2 ( x² + 5x + 6 ) + x² + 10x + 25 = 7

<=> x² + 4x + 4 - 2x² - 10x - 12 + x² + 10x + 25 =  7

<=> 4x + 17 = 7

<=> 4x = - 10

<=> x = - 5/2

Vậy x = - 5/2 là nghiệm của pt.

bạn ơi dấu giá trị tuyệt đối// à

bai nhiu qua troi

tự làm đi