\(\frac{3x}{2}=\frac{49}{30}\)

3x = 49/15

x = 49/45

b) \(\frac{3x}{2}=-\frac{61}{30}\)

3x = -61/15

x = -61/45

\(\frac{3x}{2}=\frac{49}{30}\Rightarrow3x=\frac{49.2}{30}=\frac{49}{15}\Rightarrow x=\frac{49}{45}\)

\(\frac{3x}{2}=\frac{-61}{30}\Rightarrow3x=\frac{\left(-61\right).2}{30}=\frac{-61}{15}\Rightarrow x=\frac{-61}{45}\)

a. \(\frac{3x}{2}=\frac{49}{30}\)

<=> 3x . 30 = 2 . 49

<=> 90x      = 98

<=> x          = 98 : 90

<=> x          = \(\frac{49}{45}\)

b. \(\frac{3x}{2}=-\frac{61}{30}\)

<=> 3x . 30 = - 61 . 2

<=> 90x      = -122

<=> x          = -122 : 90

<=> x          = \(-\frac{61}{45}\)

a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20

10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110

10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110

10 x X - 100 = 110

10 x X = 110 + 100

10 x X = 210

       X = 210 : 10

       X = 21

a 10 x X-1-3-5-7-....-19 = 2+4+6+....+20

​10xX-1-3-5-7-....-19=110

​10xX=110+1+3+5+7+....+19

​10xX=210

​X=210:10

​X=21

b là 4

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ĐKXĐ: \(x\ge-1\)

\(3x^2-2x-2=\dfrac{6}{\sqrt{30}}\sqrt{\left(x+1\right)\left(x^2+2x+2\right)}\)

\(\Leftrightarrow3\left(x^2+2x+2\right)-8\left(x+1\right)=\dfrac{6}{\sqrt{30}}\sqrt{\left(x+1\right)\left(x^2+2x+2\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x+2}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow3a^2-8b^2-\dfrac{6}{\sqrt{30}}ab=0\)

\(\Leftrightarrow\left(3a-\sqrt{30}b\right)\left(a+\dfrac{4\sqrt{30}}{15}b\right)=0\)

\(\Leftrightarrow3a=\sqrt{30}b\)

\(\Leftrightarrow3\sqrt{x^2+2x+2}=\sqrt{30\left(x+1\right)}\)

\(\Leftrightarrow9\left(x^2+2x+2\right)=30\left(x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=2\end{matrix}\right.\)

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

3x²-15x-(3x²-x-2)=30

3x²-15x-3x²+x+2=30

-14x+2=30

-14x=28

x=-2

Bạn lên trang wed h.vn để tham khảo nhé nếu ko ai trả lời!

hok tốt!

\(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)

\(\Leftrightarrow48x-16=30\)

\(\Leftrightarrow48x=16+30\)

\(\Leftrightarrow48x=46\)

\(\Leftrightarrow x=\frac{48}{46}\)

Đề như thế này ak???

\(2\left(3x+5\right)\times\sqrt{x^2}+9=3x+2x+30\)

Bỏ số 9 vào căn luôn bạn ạ

\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)

\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)

Tham khảo nè:

Câu hỏi của Nguyễn Đăng Khôi - Toán lớp 8 - Học toán với OnlineMath

Ko chắc đâu nha :v

\(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)

\(\Leftrightarrow\sqrt{x^2+9}=\dfrac{3x^2+2x+30}{2\left(3x+5\right)}\)

\(\Leftrightarrow\sqrt{x^2+9}-3=\dfrac{3x^2+2x+30}{2\left(3x+5\right)}-3\)

\(\Leftrightarrow\dfrac{x^2+9-9}{\sqrt{x^2+9}+3}=\dfrac{3x^2-16x}{6x+10}\)

\(\Leftrightarrow\dfrac{x^2}{\sqrt{x^2+9}+3}-\dfrac{x\left(3x-16\right)}{6x+10}=0\)

\(\Leftrightarrow x\left(\dfrac{x}{\sqrt{x^2+9}+3}-\dfrac{3x-16}{6x+10}\right)=0\)

\(\Rightarrow x=0\)