= 25/49 *( 21/29-7/29) +24/49 * 15/29

=25/49*14/29+ 24/49*15/29

   BẠN TỰ LÀM TIẾP NHA

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\(\frac{25}{49}.\frac{21}{29}-\frac{25}{49}.\frac{7}{29}+\frac{24}{49}.\frac{15}{29}\)

\(=\frac{25}{49}.\left(\frac{21}{29}-\frac{7}{29}\right)+\frac{24}{49}.\frac{15}{29}\)

\(=\frac{25}{49}.\frac{14}{29}+\frac{24}{49}.\frac{15}{29}\)

\(=\left(\frac{25}{49}+\frac{24}{49}\right).\left(\frac{14}{29}+\frac{15}{29}\right)\)

\(=1.1=1\)

a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)

\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)

\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)

nên x-128=0

hay x=128

Vậy: x=128

b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)

\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)

\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)

Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)

nên x-1999=0

hay x=1999

Vậy: x=1999

a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4

<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)

<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128

b)Tương tự

<=>x-128=0

<=>x=128

Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0

b)tương tự

Ta có:

\(4x^3-3=29\)

\(\Rightarrow4x^3=32\)

\(\Rightarrow x^3=8\)

=> x=2

Thay x=2 vào \(\frac{x+16}{9}=\frac{y-25}{-16}\) ta đuợc:

\(\frac{18}{9}=\frac{y-25}{-16}\)

=> \(2=\frac{y-25}{-16}\)

=> y-25=-32

=> y=-7

Thay y=-7 vào \(\frac{y-25}{-16}=\frac{z+49}{25}\) ta đuợc:

\(2=\frac{z+49}{25}\)

=> z+49=50

=> z=1

=> x+2y+3z=2+2.(-7)+3.1=2+(-14)+3=-12+3=-9

:) chắc còn cách khác hay hơn để mk suy nghĩ

Bùng nổ Saiya

Có :

\(4x^3-3=29\)

\(\Rightarrow4x^3=32\)

\(\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\)

\(\Rightarrow\frac{y-25}{-16}=2\)

Và \(\frac{z+49}{25}=2\)

\(\Rightarrow y-25=\left(-16\right)\cdot2=-32\)

Và \(z+49=25\cdot2=50\)

\(\Rightarrow y=-7;z=1\)

\(\Rightarrow2y=-14;z=3\)

\(\Rightarrow x+2y+3z=2+\left(-14\right)+3=-9\)

Vậy \(x+2y+3z=-9\)

= -9 nhé

Có: \(4x^3_{ }-3=29\)

=>\(4x^3=32\)

=>\(x^3=8\)

=>\(x^3=2^3\)

=>x=2

=>\(\frac{x+16}{9}=\frac{2+16}{9}=\frac{18}{9}=2\)

=>\(\frac{y-25}{-16}=2\)

và \(\frac{z+49}{25}=2\)

=>\(y_{ }-25=\left(-16\right).2=-32\)

và \(z+49=25.2=50\)

=>y=-7; z=1

=>2y=-14 ; z=3

=>x+2y+3z=2+(-14)+3=-9

Vậy x+2y+3z=-9

Ta có: 4x³-3 =29

4x³=32

x³=8

x=2

Thay x=2 vào biểu thức \(\frac{x+16}{9}\) ta được: \(\frac{2+16}{9}=\frac{18}{9}=2\)

\(\Rightarrow\frac{y-25}{-16}=2\Rightarrow y=2\cdot\left(-16\right)+25=-7\)

\(\Rightarrow\frac{z+49}{25}=2\Rightarrow z=2\cdot25-49=1\)

Vậy \(x-2y+3z=2-2\left(-7\right)+3\cdot1=2+14+3=19\)

k' nhé

ai làm cho mình đúng đều có k hết

\(4x^3-3=29\Rightarrow4x^3=32\Rightarrow x^3=8\Rightarrow x=2\)

Thay vào: \(\frac{x+16}{9}=\frac{y-25}{-16}=\frac{z+49}{25}\)

\(\Rightarrow\frac{2+16}{9}=\frac{y-25}{-16}=\frac{z+49}{25}\Rightarrow\frac{y-25}{-16}=\frac{z+49}{25}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y=2.\left(-16\right)+25=-7\\z=2.25-49=1\end{matrix}\right.\)

\(\Rightarrow x-2y+3z=2-2.\left(-7\right)+3.1=2+14+3=19\)

\(4x^3-3=29\)

\(\Rightarrow4x^3=32\Rightarrow x^3=8\Rightarrow x=2\)

\(\Rightarrow\frac{x+16}{9}=\frac{2+16}{9}=2\)

Ta có

\(\frac{x+16}{9}=\frac{y-25}{-16}\Rightarrow\frac{y-25}{-16}=2\)

\(\Rightarrow y-25=-32\Rightarrow y=-7\);

\(\frac{x+16}{9}=\frac{z+49}{25}\Rightarrow\frac{z+49}{25}=2\Rightarrow z+49=50\Rightarrow z=1\)

Vậy x = 2 ; y = -7 ; z = 1

\(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) (1)

Ta có: \(4x^3-3=29\)

\(\Rightarrow4x^3=32\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào điều (1) ta có:

\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=\dfrac{18}{9}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y-25=2.\left(-16\right)\\z+49=2.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y-25=-32\\z+49=50\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=-7\\z=1\end{matrix}\right.\)

Vậy giá trị của biểu thức \(x-2y+3z\) là:

\(2-2.\left(-7\right)+3.1=2+14+3=19\)

Tìm x:

\(\dfrac{4-x}{6-x}\)=\(\dfrac{x-3}{x-8}\)\(\Rightarrow\)(4-x)(x-8)=(6-x)(x-3)

\(\Rightarrow\)12x-x²-32=9x-x²-18

\(\Rightarrow\)3x=14\(\Rightarrow\)x=\(\dfrac{14}{3}\).

\(\dfrac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)

=\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)

=\(\dfrac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)

=5.(1-7.2²) = 5.(1-28) = 5.(-27) = -135

không ra đâu: không phải x⁴ mà là x³

VẬy chắc mình ghi đề sai . Bạn tính giúp mình \(4x^3\)vì mình nhìn nó bị mờ do chụp