3 – 4x(25 – 2x) = 8x2 + x – 300

⇔ 3 – 4x.25 + 4x.2x = 8x2 + x – 300

⇔ 3 – 100x + 8x2 = 8x2 + x – 300

⇔ -100x – x = -300 – 3

⇔ -101x = -303

⇔ x = 3.

Vậy phương trình có tập nghiệm S = {3}.

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`c,3-4x(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>101x=303`

`<=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

câu bc hình bị lỗi

a) (x - 1) - (2x - 1) = 9 - x

<=> x - 1 - 2x + 1 = 9 - x

<=> x - 2x + x = 9 + 1 - 1

<=> 0x = 9 (vô lý vì 0x = 0 với mọi x)

Vậy PT vô nghiệm

b) 3 - 4x. (25 - 2x) = 8x²

<=> 3 - 100x + 8x² = 8x²

<=> 3 - 100x = 0

<=> -100x = -3

<=> 100x = 3

\(\Leftrightarrow x=\dfrac{3}{100}\)

Vậy: \(S=\left\{\dfrac{3}{100}\right\}\)

a) (x-1) - (2x-1) = 9-x

<=> x - 1 - 2x + 1= 9-x

<=> -x = 9-x

<=> -x + x = 9 

<=> 0 = 9 ( sai )

Vậy tập nghiệm S ={\(\varnothing\)}

b) 3 - 4x(25 - 2x) = 8x²

<=> 3 - 100x + 8x² = 8x²

<=> 3 = 100x

<=> \(\dfrac{3}{100}\)= x = 0,03

Vập tập nghiệm S ={0,03}

a) (x²+x-6)(x²+9x+14) = 300

<=> (x-2)(x+3)(x+2)(x+7) - 300 = 0

<=> [(x-2)(x+7)][(x+2)(x+3)] - 300 = 0

<=> (x²-5x-14)(x²+5x+6) - 300 = 0

Đặt x² + 5x - 14 = a

<=> a(a+20) - 300 = 0

<=> a² + 20a - 300 = 0

<=> a² + 20a + 100 - 400 = 0

<=> (a+10)² - 20² = 0

<=> (a-10)(a+30) = 0

<=> \(\left[{}\begin{matrix}a=10\\a=-30\end{matrix}\right.\)

Với a = 10, ta có:

x² + 5x - 14 = 10

=> x² + 5x - 24 = 0

=> (x-3)(x+8) = 0

=> \(\left[{}\begin{matrix}x=3\\x=-8\end{matrix}\right.\)

Với a = -30, ta có:

x² + 5x - 14 = -30

=> x² + 5x + 16 = 0 (vn)

Vậy nghiệm pt x = 3; x = -8

b) (2x-5)(3x+1) = 4x² - 25

<=> (2x-5)(3x+1) = (2x-5)(2x+5)

<=> (2x-5)(3x+1-2x-5) = 0

<=> (2x-5)(x-4) = 0

<=> \(\left[{}\begin{matrix}2x-5=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=4\end{matrix}\right.\)

Vậy...

\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)

\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)

\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)

Đề câu d của bạn hình như sai dấu ý

sửa lại

.d. 3 - 4x (25 - 2x) = 8x ² + x - 300

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow-101x=-303\)

\(\Leftrightarrow x=3\)

Vậy phương trình có nghiệm \(x=3\)

Chúc bạn học tốt!

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)