Ta có : \(\frac{1+2x}{36}=\frac{1+4x}{48}=\frac{1+6x}{6y}\Rightarrow\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{1+2x}{36}=\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}=\frac{2+4x-1-4x}{72-48}=\frac{1}{24}\)

=> \(\frac{1+4x}{48}=\frac{1}{24}\Rightarrow\frac{1+4x}{48}=\frac{2}{48}\Rightarrow1+4x=2\Rightarrow x=0,25\)

\(\frac{1+2x}{36}=\frac{1+4x}{48}=\frac{1+6x}{6x}\Rightarrow\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\frac{1+2x}{36}=\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}=\frac{2+4x-1-4x}{72-48}=\frac{1}{24}\)

\(\Rightarrow\frac{1+4x}{48}=\frac{1}{24}\Rightarrow\frac{1+4x}{48}=\frac{2}{48}\Rightarrow1+4x=2\Rightarrow x=0,25\)

x⁴+4x³-4x²-48x-48=0

=> x⁴+4(x³-x²) - 48x = 48

=> x⁴ + 4[x²(x-1)] - 48x = 48 

     \(x^4+4x^3-4x^2-48x-48=0\)

\(\Leftrightarrow\)\(x^4-2x^3-4x^2+6x^3-12x^2-24x+12x^2-24x-48=0\)

\(\Leftrightarrow\)\(x^2\left(x^2-2x-4\right)+6x\left(x^2-2x-4\right)+12\left(x^2-2x-4\right)=0\)

\(\Leftrightarrow\)\(\left(x^2-2x-4\right)\left(x^2+6x+12\right)\)

\(\Leftrightarrow\)\(\left[\left(x-1\right)^2-5\right]\left(x^2+6x+12\right)=0\)

\(\Leftrightarrow\)\(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\left(x^2+6x+12\right)=0\)

Ta có:   \(x^2+6x+12=\left(x+3\right)^2+3>0\)

\(\Rightarrow\)\(\orbr{\begin{cases}x-1-\sqrt{5}=0\\x-1+\sqrt{5}=0\end{cases}}\)      

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}\)

Vậy...

`a)sqrt{1-4x+4x^2}+5=x-2`

`<=>\sqrt{(2x-1)^2}=x-2-5`

`<=>|2x-1|=x-7(x>=7)`

`<=>[(2x-1=x-7),(2x-1=7-x):}`

`<=>[(x=-6(ktm)),(3x=8):}`

`<=>x=8/3(ktm)`

Vậy PTVN

`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`

`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`

`<=>4sqrt{x+3}=4`

`<=>sqrt{x+3}=1<=>x+3=1`

`<=>x=-2(tm)`

Vậy `S={-2}`

a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)

TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)

Vậy \(S=\varnothing\)

b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)

a. \(\sqrt{1-4x+4x^2}+5=x-2\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\)

\(\Leftrightarrow\left|1-2x\right|-x=-7\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x-x=-7\\2x-1-x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\x=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-6\end{matrix}\right.\)

b. ĐKXĐ: \(x\ge-3\)
\(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\)

\(\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}-6\sqrt{3+x}=4\)

\(\Leftrightarrow4\sqrt{3+x}=4\) \(\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\) ( thỏa mãn đk )

4X=3Y =>x/3=y/4

cho x/3=y/4=k

Ta có XxY/3x4=k^2

48/12=k^2

k^2=4=>k=2

x/3=k=2=>x=2x3=6

y/4=k=2=>y=2x4=8

a)\(x^2+10x=24\)

\(\Leftrightarrow x^2+10x-24=0\)

\(\Leftrightarrow x^2-2x+12x-24=0\)

\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)

b)\(4x^2+4x=24\)

\(\Leftrightarrow4x^2+4x-24=0\)

\(\Leftrightarrow4\left(x^2+x-6\right)=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)

c)\(4x^2-4x=48\)

\(\Leftrightarrow4x^2-4x-48=0\)

\(\Leftrightarrow4\left(x^2-x-12\right)=0\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow x^2+3x-4x-12=0\)

\(\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

\(a,x^2+10x=24\)
\(\Leftrightarrow x^2+10x-24=0\)
\(\Leftrightarrow x^2-2x+12x-24=0\)
\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-12 }\)
\(b,4x^2+4x=24\)
\(\Leftrightarrow4x^2+4x-24=0\)
\(\Leftrightarrow4x^2-8x+12x-24=0\)
\(=4x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
Vậy hoặc \(\text{Vậy x=2 hoặc x=-3 }\)
\(c,4x^2-4x=48\)
\(\Leftrightarrow4x^2-4x-48=0\)
\(\Leftrightarrow\left[\left(2x\right)^2-2.2x+1^2\right]-1^2-48=0\)
\(\Leftrightarrow\left(2x-1\right)^2-49=0\)
\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-8=0\\2x+6=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(\text{Vậy x=4 hoặc x=-3 }\)

bạn chuyển vế tất cả bên phải sang bên trái rồi giải theo vi-et hoặc đưa về pt tích là OK..:)

a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50

b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1

c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10

d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13

ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0

13B 14D

Câu 13. Tìm x biết  4x + 12 = 48, ta được x bằng:

A.32                                      B.9                         C.15                       D.54

Câu 14. Tìm x biết  2346 : (x + 8) = 23, ta được x bằng:

A.110                                    B.92                       C.93                       D.94

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